Chapter 10 Straight Lines

Given:

Various conditions defining lines.

---

To Find:

The slope of each given line.

---

Solution:

We use the following formulas for the slope of a line:

1. If a line passes through two points $(x_1, y_1)$ and $(x_2, y_2)$, its slope is $m = \frac{y_2 - y_1}{x_2 - x_1}$, provided $x_1 \neq x_2$.

2. If a line makes an inclination $\theta$ with the positive direction of the x-axis, its slope is $m = \tan \theta$, provided $\theta \neq 90^\circ$.

---

(a) Passing through the points (3, – 2) and (–1, 4)

Let $(x_1, y_1) = (3, -2)$ and $(x_2, y_2) = (-1, 4)$.

Using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$:

$m = \frac{4 - (-2)}{-1 - 3}$

$m = \frac{4 + 2}{-4}$

$m = \frac{6}{-4}$

$m = -\frac{3}{2}$

The slope of the line is $-\frac{3}{2}$.

---

(b) Passing through the points (3, – 2) and (7, – 2)

Let $(x_1, y_1) = (3, -2)$ and $(x_2, y_2) = (7, -2)$.

Using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$:

$m = \frac{-2 - (-2)}{7 - 3}$

$m = \frac{-2 + 2}{4}$

$m = \frac{0}{4}$

$m = 0$

The slope of the line is $0$. This indicates a horizontal line.

---

(c) Passing through the points (3, – 2) and (3, 4)

Let $(x_1, y_1) = (3, -2)$ and $(x_2, y_2) = (3, 4)$.

Using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$:

$m = \frac{4 - (-2)}{3 - 3}$

$m = \frac{4 + 2}{0}$

$m = \frac{6}{0}$

Division by zero is undefined.

The slope of the line is undefined. This indicates a vertical line.

---

(d) Making inclination of 60° with the positive direction of x-axis

The inclination is $\theta = 60^\circ$.

Using the formula $m = \tan \theta$:

$m = \tan 60^\circ$

We know that $\tan 60^\circ = \sqrt{3}$.

$m = \sqrt{3}$

The slope of the line is $\sqrt{3}$.

Given:

Angle between two lines, $\theta = \frac{\pi}{4}$.

Slope of one line, $m_1 = \frac{1}{2}$.

---

To Find:

The slope of the other line (let's call it $m_2$).

---

Solution:

The formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by:

$\tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|$

We are given $\theta = \frac{\pi}{4}$ and $m_1 = \frac{1}{2}$. We know that $\tan \frac{\pi}{4} = 1$.

Substitute the given values into the formula:

$1 = \left|\frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2} m_2}\right|$

Simplify the expression inside the absolute value:

$1 = \left|\frac{\frac{2m_2 - 1}{2}}{\frac{2 + m_2}{2}}\right|$

$1 = \left|\frac{2m_2 - 1}{2 + m_2}\right|$

This equation implies two possible cases:

Case 1: $\frac{2m_2 - 1}{2 + m_2} = 1$

Assuming $2 + m_2 \neq 0$, multiply both sides by $(2 + m_2)$:

$2m_2 - 1 = 2 + m_2$

$2m_2 - m_2 = 2 + 1$

$m_2 = 3$

Check $1 + m_1 m_2 = 1 + \frac{1}{2}(3) = 1 + \frac{3}{2} = \frac{5}{2} \neq 0$. This solution is valid.

Case 2: $\frac{2m_2 - 1}{2 + m_2} = -1$

Assuming $2 + m_2 \neq 0$, multiply both sides by $(2 + m_2)$:

$2m_2 - 1 = -(2 + m_2)$

$2m_2 - 1 = -2 - m_2$

$2m_2 + m_2 = -2 + 1$

$3m_2 = -1$

$m_2 = -\frac{1}{3}$

Check $1 + m_1 m_2 = 1 + \frac{1}{2}(-\frac{1}{3}) = 1 - \frac{1}{6} = \frac{5}{6} \neq 0$. This solution is also valid.

Thus, there are two possible values for the slope of the other line.

The slope of the other line is $3$ or $-\frac{1}{3}$.

Given:

Line 1 passes through points $(-2, 6)$ and $(4, 8)$.

Line 2 passes through points $(8, 12)$ and $(x, 24)$.

Line 1 is perpendicular to Line 2.

---

To Find:

The value of $x$.

---

Solution:

The slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ (provided $x_1 \neq x_2$).

Let $m_1$ be the slope of the line passing through $(-2, 6)$ and $(4, 8)$.

$m_1 = \frac{8 - 6}{4 - (-2)}$

$m_1 = \frac{2}{4 + 2}$

$m_1 = \frac{2}{6}$

$m_1 = \frac{1}{3}$

Let $m_2$ be the slope of the line passing through $(8, 12)$ and $(x, 24)$.

$m_2 = \frac{24 - 12}{x - 8}$

$m_2 = \frac{12}{x - 8}$

Given that the two lines are perpendicular, the product of their slopes is $-1$, provided neither line is vertical or horizontal.

Since $m_1 = \frac{1}{3}$ is a finite non-zero slope, neither line is vertical or horizontal. Thus, their slopes must satisfy the condition $m_1 m_2 = -1$.

Substitute the expressions for $m_1$ and $m_2$:

$\frac{1}{3} \times \frac{12}{x - 8} = -1$

$\frac{12}{3(x - 8)} = -1$

$\frac{4}{x - 8} = -1$

Assuming $x - 8 \neq 0$, multiply both sides by $(x - 8)$:

$4 = -1 \times (x - 8)$

$4 = -x + 8$

Add $x$ to both sides and subtract 4 from both sides:

$x = 8 - 4$

$x = 4$

The value $x=4$ ensures $x-8 = 4-8 = -4 \neq 0$, so the slope $m_2$ is defined.

The value of $x$ is 4.

Given:

Three points P$(h, k)$, Q$(x_1, y_1)$, and R$(x_2, y_2)$ lie on a line (are collinear).

---

To Show:

$(h - x_1) (y_2 - y_1) = (k - y_1) (x_2 - x_1)$.

---

Proof:

Since the points P, Q, and R are collinear, the slope of the line segment PQ must be equal to the slope of the line segment QR (provided the slopes are defined).

The slope of the line passing through two points $(x_a, y_a)$ and $(x_b, y_b)$ is given by $m = \frac{y_b - y_a}{x_b - x_a}$, provided $x_a \neq x_b$.

Slope of PQ ($m_{PQ}$) using points P$(h, k)$ and Q$(x_1, y_1)$:

$m_{PQ} = \frac{k - y_1}{h - x_1}$ (assuming $h \neq x_1$)

Slope of QR ($m_{QR}$) using points Q$(x_1, y_1)$ and R$(x_2, y_2)$:

$m_{QR} = \frac{y_2 - y_1}{x_2 - x_1}$ (assuming $x_1 \neq x_2$)

Since P, Q, R are collinear, $m_{PQ} = m_{QR}$ (if the line is not vertical).

$\frac{k - y_1}{h - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$

Cross-multiply the terms (assuming $h \neq x_1$ and $x_2 \neq x_1$):

$(k - y_1)(x_2 - x_1) = (h - x_1)(y_2 - y_1)$

This is the required equation, just with sides swapped:

$(h - x_1)(y_2 - y_1) = (k - y_1)(x_2 - x_1)$

Consider the special cases:

Case 1: Vertical line

If the three points lie on a vertical line, their x-coordinates are the same. Thus, $h = x_1 = x_2$.

In this case, $h - x_1 = 0$ and $x_2 - x_1 = 0$.

The equation to be shown becomes $(0)(y_2 - y_1) = (k - y_1)(0)$, which simplifies to $0 = 0$. This is true.

Case 2: Horizontal line

If the three points lie on a horizontal line, their y-coordinates are the same. Thus, $k = y_1 = y_2$.

In this case, $k - y_1 = 0$ and $y_2 - y_1 = 0$.

The equation to be shown becomes $(h - x_1)(0) = (0)(x_2 - x_1)$, which simplifies to $0 = 0$. This is true.

In all cases (non-vertical, non-horizontal, vertical, or horizontal), the relationship $(h - x_1)(y_2 - y_1) = (k - y_1)(x_2 - x_1)$ holds true if the points P, Q, and R are collinear.

Hence, shown.

Given:

The time and distance graph of a linear motion passes through two points recorded as:

When $T = 0$, $D = 2$. This corresponds to the point $(T_1, D_1) = (0, 2)$.

When $T = 3$, $D = 8$. This corresponds to the point $(T_2, D_2) = (3, 8)$.

---

To Find:

The law of motion, which describes how distance (D) depends upon time (T), using the concept of slope.

---

Solution:

Since the time and distance graph represents a linear motion, it is a straight line. The relationship between distance (D) and time (T) can be expressed in the form of a linear equation: $D = mT + c$, where $m$ is the slope of the line and $c$ is the D-intercept (the value of D when $T=0$).

The slope ($m$) of the line passing through two points $(T_1, D_1)$ and $(T_2, D_2)$ is given by the formula:

$m = \frac{D_2 - D_1}{T_2 - T_1}$

Using the given points $(0, 2)$ and $(3, 8)$:

$m = \frac{8 - 2}{3 - 0}$

$m = \frac{6}{3}$

$m = 2$

The D-intercept ($c$) is the value of D when $T = 0$. From the given information, when $T = 0$, $D = 2$.

$c = 2$

Substitute the values of the slope ($m=2$) and the D-intercept ($c=2$) into the linear equation $D = mT + c$:

$D = 2T + 2$

This equation represents the law of motion, showing how the distance D depends on time T.

The law of motion is $D = 2T + 2$.

Given:

The vertices of a quadrilateral are A(– 4, 5), B(0, 7), C(5, – 5) and D(– 4, –2).

---

To Do:

1. Draw the quadrilateral in the Cartesian plane.

2. Find the area of the quadrilateral.

---

Solution:

1. Drawing the Quadrilateral

The vertices A, B, C, and D are plotted on a Cartesian plane and joined in order to form the quadrilateral ABCD.

---

2. Finding the Area of the Quadrilateral

To find the area of the quadrilateral ABCD, we can divide it into two triangles, $\triangle ABC$ and $\triangle ADC$, by drawing the diagonal AC.

Area(ABCD) = Area($\triangle ABC$) + Area($\triangle ADC$)

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

Area = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Area of $\triangle ABC$:

The vertices are A(– 4, 5), B(0, 7), and C(5, – 5).

Area($\triangle ABC$) = $\frac{1}{2} |(-4)(7 - (-5)) + 0(-5 - 5) + 5(5 - 7)|$

Area($\triangle ABC$) = $\frac{1}{2} |(-4)(12) + 0(-10) + 5(-2)|$

Area($\triangle ABC$) = $\frac{1}{2} |-48 + 0 - 10|$

Area($\triangle ABC$) = $\frac{1}{2} |-58| = \frac{58}{2} = 29$ square units.

Area of $\triangle ADC$:

The vertices are A(– 4, 5), D(– 4, –2), and C(5, – 5).

Area($\triangle ADC$) = $\frac{1}{2} |(-4)(-2 - (-5)) + (-4)(-5 - 5) + 5(5 - (-2))|$

Area($\triangle ADC$) = $\frac{1}{2} |(-4)(3) + (-4)(-10) + 5(7)|$

Area($\triangle ADC$) = $\frac{1}{2} |-12 + 40 + 35|$

Area($\triangle ADC$) = $\frac{1}{2} |63| = \frac{63}{2} = 31.5$ square units.

Total Area of Quadrilateral ABCD:

Area(ABCD) = Area($\triangle ABC$) + Area($\triangle ADC$)

Area(ABCD) = $29 + 31.5 = 60.5$ square units.

---

Alternate Solution (Using Shoelace Formula):

The area of a polygon with vertices $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$ listed in order (counterclockwise or clockwise) is given by:

Area = $\frac{1}{2} |(x_1y_2 + x_2y_3 + \dots + x_ny_1) - (y_1x_2 + y_2x_3 + \dots + y_nx_1)|$

Using vertices A(– 4, 5), B(0, 7), C(5, – 5), and D(– 4, –2):

Area = $\frac{1}{2} |[(-4)(7) + (0)(-5) + (5)(-2) + (-4)(5)] \ $$ - [(5)(0) + (7)(5) \ $$ + (-5)(-4) + (-2)(-4)]|$

Area = $\frac{1}{2} |[-28 + 0 - 10 - 20] - [0 + 35 + 20 + 8]|$

Area = $\frac{1}{2} |[-58] - [63]|$

Area = $\frac{1}{2} |-58 - 63|$

Area = $\frac{1}{2} |-121| = \frac{121}{2} = 60.5$ square units.

---

The area of the quadrilateral is 60.5 square units.

Given:

An equilateral triangle with side length $2a$.

The base lies along the y-axis.

The midpoint of the base is at the origin O(0, 0).

---

To Find:

The coordinates of the three vertices of the triangle.

---

Solution:

Let the equilateral triangle be denoted as $\triangle ABC$.

Let the base of the triangle be BC. The length of the base is $BC = 2a$.

Since the base BC lies on the y-axis and its midpoint is the origin (0, 0), the two endpoints of the base will be at a distance of 'a' from the origin along the y-axis.

So, the coordinates of the vertices B and C are B(0, a) and C(0, -a).

Let the third vertex be A with coordinates $(x, y)$.

In an equilateral triangle, the altitude from a vertex to the opposite side bisects the side. The altitude from vertex A to the base BC will lie along the x-axis, as the x-axis is the perpendicular bisector of the base BC.

Therefore, the y-coordinate of vertex A is 0. So, A has coordinates $(x, 0)$.

The length of the side AB is also $2a$. We can use the distance formula to find the value of x.

Distance $AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

$AB^2 = (2a)^2 = 4a^2$

Using points A(x, 0) and B(0, a):

$AB^2 = (x-0)^2 + (0-a)^2$

$4a^2 = x^2 + (-a)^2$

$4a^2 = x^2 + a^2$

$x^2 = 4a^2 - a^2$

$x^2 = 3a^2$

$x = \pm \sqrt{3}a$

This means there are two possible positions for vertex A, either on the positive x-axis or the negative x-axis.

So, the coordinates of vertex A can be $(a\sqrt{3}, 0)$ or $(-a\sqrt{3}, 0)$.

---

Thus, there are two possible triangles. The vertices are:

Case 1: A($a\sqrt{3}$, 0), B(0, a), and C(0, -a).

Case 2: A(-$a\sqrt{3}$, 0), B(0, a), and C(0, -a).

Given:

Two points P$(x_1, y_1)$ and Q$(x_2, y_2)$.

---

To Find:

The distance between P and Q when:

(i) The line segment PQ is parallel to the y-axis.

(ii) The line segment PQ is parallel to the x-axis.

---

Solution:

The general formula for the distance between two points P$(x_1, y_1)$ and Q$(x_2, y_2)$ is:

(i) When PQ is parallel to the y-axis

If a line is parallel to the y-axis, all points on the line have the same x-coordinate. Therefore, for the points P and Q, we must have:

$x_1 = x_2$

This means $x_2 - x_1 = 0$.

Substituting this condition into the distance formula (i):

$PQ = \sqrt{(0)^2 + (y_2 - y_1)^2}$

$PQ = \sqrt{(y_2 - y_1)^2}$

Since the distance must be a non-negative value, we take the absolute value:

$PQ = |y_2 - y_1|$

---

(ii) When PQ is parallel to the x-axis

If a line is parallel to the x-axis, all points on the line have the same y-coordinate. Therefore, for the points P and Q, we must have:

$y_1 = y_2$

This means $y_2 - y_1 = 0$.

Substituting this condition into the distance formula (i):

$PQ = \sqrt{(x_2 - x_1)^2 + (0)^2}$

$PQ = \sqrt{(x_2 - x_1)^2}$

Since the distance must be a non-negative value, we take the absolute value:

$PQ = |x_2 - x_1|$

Given:

Two points, A(7, 6) and B(3, 4).

---

To Find:

The coordinates of a point on the x-axis that is equidistant from points A and B.

---

Solution:

Let the required point on the x-axis be P. Since any point on the x-axis has a y-coordinate of 0, we can denote its coordinates as P(x, 0).

The problem states that P is equidistant from A and B. This means the distance PA is equal to the distance PB.

$PA = PB$

To avoid dealing with square roots, we can square both sides:

$PA^2 = PB^2$

Using the distance formula, $(d^2) = (x_2 - x_1)^2 + (y_2 - y_1)^2$, we can write expressions for $PA^2$ and $PB^2$.

For $PA^2$, using P(x, 0) and A(7, 6):

$PA^2 = (7 - x)^2 + (6 - 0)^2 = (7 - x)^2 + 36$

For $PB^2$, using P(x, 0) and B(3, 4):

$PB^2 = (3 - x)^2 + (4 - 0)^2 = (3 - x)^2 + 16$

Now, we set $PA^2 = PB^2$:

Expand the squared terms:

$(49 - 14x + x^2) + 36 = (9 - 6x + x^2) + 16$

$x^2 - 14x + 85 = x^2 - 6x + 25$

Subtract $x^2$ from both sides:

$-14x + 85 = -6x + 25$

Rearrange the terms to solve for x:

$85 - 25 = -6x + 14x$

$60 = 8x$

$x = \frac{60}{8} = \frac{15}{2}$

So, the x-coordinate of the point P is $\frac{15}{2}$. The y-coordinate is 0.

---

The required point on the x-axis is $\left(\frac{15}{2}, 0\right)$.

Given:

A line passes through the origin O(0, 0).

The same line also passes through the midpoint of the segment joining P(0, –4) and B(8, 0).

---

To Find:

The slope of the line.

---

Solution:

Step 1: Find the coordinates of the midpoint.

Let M be the midpoint of the line segment joining P(0, –4) and B(8, 0).

The midpoint formula is $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.

Substituting the coordinates of P and B:

$M = \left(\frac{0 + 8}{2}, \frac{-4 + 0}{2}\right)$

$M = \left(\frac{8}{2}, \frac{-4}{2}\right)$

$M = (4, -2)$

---

Step 2: Find the slope of the line passing through the origin and the midpoint.

The line passes through O(0, 0) and M(4, -2).

The formula for the slope (m) of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Substituting the coordinates of O and M:

$m = \frac{-2 - 0}{4 - 0}$

$m = \frac{-2}{4}$

$m = -\frac{1}{2}$

---

The slope of the line is $-\frac{1}{2}$.

Given:

Three points A(4, 4), B(3, 5), and C(–1, –1).

---

To Show:

The points A, B, and C are the vertices of a right-angled triangle, without using the Pythagoras theorem.

---

Proof:

To prove that the triangle is right-angled without using the Pythagoras theorem, we can use the property of slopes of perpendicular lines. Two non-vertical lines are perpendicular if and only if the product of their slopes is -1.

We will calculate the slopes of the three sides of the triangle: AB, BC, and AC.

The formula for the slope (m) of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Slope of side AB ($m_{AB}$):

Using points A(4, 4) and B(3, 5):

$m_{AB} = \frac{5 - 4}{3 - 4} = \frac{1}{-1} = -1$

Slope of side BC ($m_{BC}$):

Using points B(3, 5) and C(–1, –1):

$m_{BC} = \frac{-1 - 5}{-1 - 3} = \frac{-6}{-4} = \frac{3}{2}$

Slope of side AC ($m_{AC}$):

Using points A(4, 4) and C(–1, –1):

$m_{AC} = \frac{-1 - 4}{-1 - 4} = \frac{-5}{-5} = 1$

Now, we check the product of the slopes to see if any pair is perpendicular.

$m_{AB} \times m_{AC} = (-1) \times (1) = -1$

Since the product of the slopes of sides AB and AC is -1, the side AB is perpendicular to the side AC.

This means the angle at vertex A, $\angle BAC$, is $90^\circ$.

Therefore, the triangle formed by the points A, B, and C is a right-angled triangle.

---

Hence, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.

Given:

A line makes an angle of 30° with the positive direction of the y-axis, measured in the anticlockwise direction.

---

To Find:

The slope of the line.

---

Solution:

The slope ($m$) of a line is defined as the tangent of its inclination, $\theta$, where $\theta$ is the angle the line makes with the positive direction of the x-axis, measured anticlockwise.

First, we need to find the inclination $\theta$.

The positive direction of the y-axis itself makes an angle of $90^\circ$ with the positive direction of the x-axis (measured anticlockwise).

The given line makes an additional angle of $30^\circ$ with the positive y-axis, also measured anticlockwise.

Therefore, the total angle of the line from the positive x-axis (the inclination $\theta$) is the sum of these two angles:

$\theta = 90^\circ + 30^\circ = 120^\circ$

Now, we can calculate the slope:

$m = \tan(120^\circ)$

To evaluate $\tan(120^\circ)$, we can use the identity $\tan(180^\circ - \alpha) = -\tan(\alpha)$.

$m = \tan(180^\circ - 60^\circ)$

$m = -\tan(60^\circ)$

Since $\tan(60^\circ) = \sqrt{3}$, we have:

$m = -\sqrt{3}$

---

The slope of the line is $-\sqrt{3}$.

Given:

Three points A(x, –1), B(2, 1), and C(4, 5) are collinear.

---

To Find:

The value of x.

---

Solution:

If three points are collinear, they lie on the same straight line. This means the slope between any two pairs of these points must be the same.

We will set the slope of the line segment AB equal to the slope of the line segment BC.

Slope of AB = Slope of BC

The formula for the slope (m) between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Slope of AB: using A(x, –1) and B(2, 1)

$m_{AB} = \frac{1 - (-1)}{2 - x} = \frac{2}{2 - x}$

Slope of BC: using B(2, 1) and C(4, 5)

$m_{BC} = \frac{5 - 1}{4 - 2} = \frac{4}{2} = 2$

Now, we set the slopes equal to each other:

Multiply both sides by $(2-x)$:

$2 = 2(2 - x)$

Divide both sides by 2:

$1 = 2 - x$

Solve for x:

$x = 2 - 1$

$x = 1$

---

Alternate Solution (Using Area of Triangle):

If three points are collinear, the area of the triangle formed by these points is zero.

Area = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$

Using A(x, –1), B(2, 1), and C(4, 5):

$x(1 - 5) + 2(5 - (-1)) + 4(-1 - 1) = 0$

$x(-4) + 2(6) + 4(-2) = 0$

$-4x + 12 - 8 = 0$

$-4x + 4 = 0$

$4 = 4x$

$x = 1$

---

The value of x is 1.

Given:

Four points A(–2, –1), B(4, 0), C(3, 3), and D(–3, 2).

---

To Show:

The points A, B, C, and D are the vertices of a parallelogram, without using the distance formula.

---

Proof:

A quadrilateral is a parallelogram if its opposite sides are parallel. We can prove this by showing that the slopes of opposite sides are equal.

The formula for the slope (m) between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Step 1: Calculate the slopes of all four sides.

Slope of AB ($m_{AB}$): using A(–2, –1) and B(4, 0)

$m_{AB} = \frac{0 - (-1)}{4 - (-2)} = \frac{1}{6}$

Slope of BC ($m_{BC}$): using B(4, 0) and C(3, 3)

$m_{BC} = \frac{3 - 0}{3 - 4} = \frac{3}{-1} = -3$

Slope of CD ($m_{CD}$): using C(3, 3) and D(–3, 2)

$m_{CD} = \frac{2 - 3}{-3 - 3} = \frac{-1}{-6} = \frac{1}{6}$

Slope of DA ($m_{DA}$): using D(–3, 2) and A(–2, –1)

$m_{DA} = \frac{-1 - 2}{-2 - (-3)} = \frac{-3}{1} = -3$

Step 2: Compare the slopes of opposite sides.

Comparing the slopes of AB and CD:

$m_{AB} = \frac{1}{6}$ and $m_{CD} = \frac{1}{6}$. Since $m_{AB} = m_{CD}$, side AB is parallel to side CD.

Comparing the slopes of BC and DA:

$m_{BC} = -3$ and $m_{DA} = -3$. Since $m_{BC} = m_{DA}$, side BC is parallel to side DA.

Since both pairs of opposite sides are parallel, the quadrilateral ABCD is a parallelogram.

---

Alternate Proof (Using Midpoint of Diagonals):

A quadrilateral is a parallelogram if its diagonals bisect each other, which means they have the same midpoint.

Midpoint of diagonal AC: using A(–2, –1) and C(3, 3)

Midpoint$_{AC} = \left(\frac{-2 + 3}{2}, \frac{-1 + 3}{2}\right) = \left(\frac{1}{2}, 1\right)$

Midpoint of diagonal BD: using B(4, 0) and D(–3, 2)

Midpoint$_{BD} = \left(\frac{4 + (-3)}{2}, \frac{0 + 2}{2}\right) = \left(\frac{1}{2}, 1\right)$

Since the midpoint of AC is the same as the midpoint of BD, the diagonals bisect each other.

Hence, the given points are the vertices of a parallelogram.

Given:

A line joining the points A(3, –1) and B(4, –2).

---

To Find:

The angle between the x-axis and the line AB.

---

Solution:

The angle between a line and the positive direction of the x-axis is called the inclination of the line, denoted by $\theta$. The slope ($m$) of the line is related to the inclination by the formula:

Step 1: Find the slope of the line.

Using the points A(3, –1) and B(4, –2), the slope is:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - (-1)}{4 - 3}$

$m = \frac{-2 + 1}{1} = \frac{-1}{1} = -1$

Step 2: Find the angle from the slope.

We have $\tan \theta = -1$.

We need to find the angle $\theta$ in the range $[0^\circ, 180^\circ)$ for which the tangent is -1.

The tangent function is negative in the second quadrant. The reference angle for which $\tan \alpha = 1$ is $\alpha = 45^\circ$.

In the second quadrant, the angle is given by $\theta = 180^\circ - \alpha$.

$\theta = 180^\circ - 45^\circ = 135^\circ$

The angle is measured counter-clockwise from the positive x-axis.

---

The angle between the x-axis and the line joining the points is $135^\circ$.

Given:

Let the slopes of the two lines be $m_1$ and $m_2$.

Condition on slopes: $m_1 = 2m_2$.

The tangent of the angle $\theta$ between the lines is $\tan \theta = \frac{1}{3}$.

---

To Find:

The possible values for the slopes $m_1$ and $m_2$.

---

Solution:

The formula for the tangent of the angle between two lines is:

$\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$

Substitute the given information into the formula:

$\frac{1}{3} = \left|\frac{2m_2 - m_2}{1 + (2m_2)(m_2)}\right|$

$\frac{1}{3} = \left|\frac{m_2}{1 + 2m_2^2}\right|$

This absolute value equation leads to two possible cases:

Case 1: $\frac{m_2}{1 + 2m_2^2} = \frac{1}{3}$

$3m_2 = 1 + 2m_2^2$

$2m_2^2 - 3m_2 + 1 = 0$

Factoring the quadratic equation:

$(2m_2 - 1)(m_2 - 1) = 0$

This gives two solutions: $m_2 = \frac{1}{2}$ or $m_2 = 1$.

If $m_2 = \frac{1}{2}$, then $m_1 = 2m_2 = 1$. So, one pair of slopes is $(\frac{1}{2}, 1)$.

If $m_2 = 1$, then $m_1 = 2m_2 = 2$. So, another pair is $(1, 2)$.

---

Case 2: $\frac{m_2}{1 + 2m_2^2} = -\frac{1}{3}$

$-3m_2 = 1 + 2m_2^2$

$2m_2^2 + 3m_2 + 1 = 0$

Factoring the quadratic equation:

$(2m_2 + 1)(m_2 + 1) = 0$

This gives two solutions: $m_2 = -\frac{1}{2}$ or $m_2 = -1$.

If $m_2 = -\frac{1}{2}$, then $m_1 = 2m_2 = -1$. So, a third pair is $(-\frac{1}{2}, -1)$.

If $m_2 = -1$, then $m_1 = 2m_2 = -2$. So, a fourth pair is $(-1, -2)$.

---

The possible slopes of the lines are the pairs: $(1, 2)$, $(\frac{1}{2}, 1)$, $(-1, -2)$, and $(-\frac{1}{2}, -1)$.

Given:

A line passes through two points, P$_1(x_1, y_1)$ and P$_2(h, k)$.

The slope of the line is $m$.

---

To Show:

$k – y_1 = m (h – x_1)$

---

Proof:

By the definition of the slope of a line, the slope $m$ of a line passing through two distinct points $(x_a, y_a)$ and $(x_b, y_b)$ is given by the formula:

$m = \frac{\text{change in y}}{\text{change in x}} = \frac{y_b - y_a}{x_b - x_a}$

In our case, the two points are $(x_1, y_1)$ and $(h, k)$. Applying the slope formula to these points, we get:

This definition is valid as long as the line is not vertical, i.e., $h \neq x_1$. If the line is vertical, its slope $m$ is undefined.

To arrive at the desired form, we can multiply both sides of equation (i) by the denominator $(h - x_1)$:

$m \times (h - x_1) = \left(\frac{k - y_1}{h - x_1}\right) \times (h - x_1)$

$m(h - x_1) = k - y_1$

Rearranging the equation gives:

$k - y_1 = m(h - x_1)$

This is the point-slope form of the equation of a line.

---

Hence, it is shown that $k – y_1 = m (h – x_1)$.

Given:

Three points A(h, 0), B(a, b), and C(0, k) lie on the same line (are collinear).

It is implied that the line does not pass through the origin and has distinct intercepts, so $h \neq 0$ and $k \neq 0$.

---

To Show:

$\frac{a}{h} + \frac{b}{k} = 1$

---

Proof:

Method 1: Using Slopes

Since the three points are collinear, the slope of the segment AB must be equal to the slope of the segment BC.

Slope of AB ($m_{AB}$) using A(h, 0) and B(a, b):

$m_{AB} = \frac{b - 0}{a - h} = \frac{b}{a - h}$

Slope of BC ($m_{BC}$) using B(a, b) and C(0, k):

$m_{BC} = \frac{k - b}{0 - a} = \frac{k - b}{-a}$

Set the slopes equal: $m_{AB} = m_{BC}$

$\frac{b}{a - h} = \frac{k - b}{-a}$

Cross-multiply:

$-a(b) = (k - b)(a - h)$

$-ab = ka - kh - ab + bh$

Add $ab$ to both sides:

$0 = ka - kh + bh$

$kh = ka + bh$

Since $h \neq 0$ and $k \neq 0$, we can divide the entire equation by $hk$:

$\frac{kh}{hk} = \frac{ka}{hk} + \frac{bh}{hk}$

$1 = \frac{a}{h} + \frac{b}{k}$

Rearranging, we get $\frac{a}{h} + \frac{b}{k} = 1$.

---

Method 2: Using Equation of a Line

The points A(h, 0) and C(0, k) are the x-intercept and y-intercept of the line, respectively.

The equation of a line in intercept form is:

$\frac{x}{h} + \frac{y}{k} = 1$

Since the point B(a, b) also lies on this line, its coordinates must satisfy the equation of the line.

Substitute $x=a$ and $y=b$ into the equation:

$\frac{a}{h} + \frac{b}{k} = 1$

Hence, it is shown that $\frac{a}{h} + \frac{b}{k} = 1$.

Given:

From the graph, we have two points on the line AB representing the population in different years.

Point A = $(1985, 92)$, which means in the year 1985, the population was 92 crores.

Point B = $(1995, 97)$, which means in the year 1995, the population was 97 crores.

---

To Find:

1. The slope of the line AB.

2. The population in the year 2010.

---

Solution:

1. Slope of the line AB

Let the year be represented by the x-coordinate and the population (in crores) by the y-coordinate.

The slope ($m$) of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

Using points A$(1985, 92)$ and B$(1995, 97)$:

$x_1 = 1985, y_1 = 92$

$x_2 = 1995, y_2 = 97$

$m = \frac{97 - 92}{1995 - 1985} = \frac{5}{10} = \frac{1}{2}$

The slope of the line AB is $\frac{1}{2}$ or $0.5$. This means the population increases by 1 crore every 2 years, or 0.5 crores per year.

---

2. Population in the year 2010

Since A, B, and the point for the year 2010 lie on the same straight line, we can use the equation of the line to find the population in 2010.

The point-slope form of the equation of a line is:

$y - y_1 = m(x - x_1)$

Using the slope $m = \frac{1}{2}$ and point A$(1985, 92)$:

We want to find the population ($y$) in the year 2010 ($x$). So, we substitute $x = 2010$ into the equation (i).

$y - 92 = \frac{1}{2}(2010 - 1985)$

$y - 92 = \frac{1}{2}(25)$

$y - 92 = 12.5$

$y = 92 + 12.5$

$y = 104.5$

Since the y-axis represents the population in crores, the population in the year 2010 will be 104.5 crores.

---

The slope of the line AB is $\frac{1}{2}$, and the population in the year 2010 will be 104.5 crores.

Given:

A point in the coordinate plane is P(–2, 3).

---

To Find:

The equations of the lines that pass through P(–2, 3) and are parallel to the coordinate axes.

---

Solution:

We need to find the equations for two separate lines.

---

1. Equation of the line parallel to the x-axis:

A line parallel to the x-axis is a horizontal line. For any horizontal line, the y-coordinate of every point on the line is constant.

Since the line passes through the point (–2, 3), the constant y-coordinate for all points on this line must be 3.

Therefore, the equation of the line parallel to the x-axis is:

$y = 3$

---

2. Equation of the line parallel to the y-axis:

A line parallel to the y-axis is a vertical line. For any vertical line, the x-coordinate of every point on the line is constant.

Since the line passes through the point (–2, 3), the constant x-coordinate for all points on this line must be –2.

Therefore, the equation of the line parallel to the y-axis is:

$x = -2$

---

Thus, the equations of the required lines are $y = 3$ and $x = -2$.

Given:

The line passes through the point $(x_1, y_1) = (–2, 3)$.

The slope of the line is $m = –4$.

---

To Find:

The equation of the line.

---

Solution:

We can find the equation of the line using the point-slope form, which is given by:

Substitute the given values of the point $(x_1, y_1) = (–2, 3)$ and the slope $m = –4$ into the formula:

$y - 3 = -4(x - (-2))$

$y - 3 = -4(x + 2)$

Now, we simplify the equation:

$y - 3 = -4x - 8$

To express the equation in the general form $Ax + By + C = 0$, we move all terms to one side:

$4x + y + 3 + 8 = 0$

$4x + y + 5 = 0$

---

Alternate Solution:

We can use the slope-intercept form of the equation of a line, which is $y = mx + c$, where c is the y-intercept.

We are given the slope $m = -4$. So the equation becomes:

$y = -4x + c$

Since the line passes through the point $(–2, 3)$, these coordinates must satisfy the equation. Substitute $x = -2$ and $y = 3$ to find $c$:

$3 = -4(-2) + c$

$3 = 8 + c$

$c = 3 - 8$

$c = -5$

Now, substitute the value of $c$ back into the slope-intercept form:

$y = -4x - 5$

Rearranging this to the general form, we get:

$4x + y + 5 = 0$

---

The required equation of the line is $4x + y + 5 = 0$.

Given:

The line passes through two points, let them be $P_1(x_1, y_1) = (1, –1)$ and $P_2(x_2, y_2) = (3, 5)$.

---

To Find:

The equation of the line passing through the given points.

---

Solution:

Method 1: Using Two-Point Form

The equation of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the two-point form:

$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$

Substitute the coordinates of the given points:

$\frac{y - (-1)}{5 - (-1)} = \frac{x - 1}{3 - 1}$

$\frac{y + 1}{6} = \frac{x - 1}{2}$

To simplify, we can cross-multiply:

$2(y + 1) = 6(x - 1)$

$2y + 2 = 6x - 6$

Now, rearrange the terms to the general form $Ax + By + C = 0$:

$6x - 2y - 6 - 2 = 0$

$6x - 2y - 8 = 0$

We can divide the entire equation by 2 to simplify it:

$3x - y - 4 = 0$

---

Alternate Solution:

Method 2: Using Slope and Point-Slope Form

First, we calculate the slope ($m$) of the line using the two given points:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - (-1)}{3 - 1} = \frac{6}{2} = 3$

Now, use the point-slope form $y - y_1 = m(x - x_1)$ with the slope $m = 3$ and one of the points, say $(1, –1)$:

$y - (-1) = 3(x - 1)$

$y + 1 = 3x - 3$

Rearranging to the general form:

$3x - y - 4 = 0$

(Note: Using the other point (3, 5) would yield the same result: $y - 5 = 3(x - 3) \implies y - 5 = 3x - 9 \implies 3x - y - 4 = 0$)

---

The required equation of the line is $3x - y - 4 = 0$.

Given:

The inclination of the line is $\theta$, such that $\tan \theta = \frac{1}{2}$.

The slope ($m$) of the line is given by $m = \tan \theta$.

Therefore, the slope of the line is $m = \frac{1}{2}$.

---

To Find:

The equation of the line for two separate conditions.

---

Solution:

---

(i) y-intercept is $-\frac{3}{2}$

We are given the slope $m = \frac{1}{2}$ and the y-intercept $c = -\frac{3}{2}$.

We use the slope-intercept form of the equation of a line: $y = mx + c$.

Substitute the given values:

$y = \frac{1}{2}x + \left(-\frac{3}{2}\right)$

$y = \frac{x - 3}{2}$

Multiply both sides by 2:

$2y = x - 3$

Rearranging to the general form $Ax + By + C = 0$ gives:

$x - 2y - 3 = 0$

---

(ii) x-intercept is 4

An x-intercept of 4 means the line passes through the point $(4, 0)$.

We have the slope $m = \frac{1}{2}$ and a point on the line $(x_1, y_1) = (4, 0)$.

We use the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$.

Substitute the values:

$y - 0 = \frac{1}{2}(x - 4)$

$y = \frac{x - 4}{2}$

Multiply both sides by 2:

$2y = x - 4$

Rearranging to the general form gives:

$x - 2y - 4 = 0$

Given:

The x-intercept of the line is $a = –3$.

The y-intercept of the line is $b = 2$.

---

To Find:

The equation of the line.

---

Solution:

The intercept form of the equation of a line is given by:

$\frac{x}{a} + \frac{y}{b} = 1$

where $a$ is the x-intercept and $b$ is the y-intercept.

Substitute the given values $a = –3$ and $b = 2$ into the formula:

$\frac{x}{-3} + \frac{y}{2} = 1$

To clear the denominators, we can multiply the entire equation by the least common multiple (LCM) of 3 and 2, which is 6.

$6 \left( \frac{x}{-3} \right) + 6 \left( \frac{y}{2} \right) = 6(1)$

$-2x + 3y = 6$

To write the equation in the standard form $Ax + By + C = 0$, we move all terms to one side:

$2x - 3y + 6 = 0$

---

Alternate Solution:

The x-intercept of –3 means the line passes through the point $(–3, 0)$.

The y-intercept of 2 means the line passes through the point $(0, 2)$.

First, we find the slope ($m$) of the line using these two points:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{0 - (-3)} = \frac{2}{3}$

Now, using the slope-intercept form $y = mx + c$, where $m = \frac{2}{3}$ and the y-intercept $c = 2$:

$y = \frac{2}{3}x + 2$

Multiply the entire equation by 3 to eliminate the fraction:

$3y = 2x + 6$

Rearranging to the standard form gives:

$2x - 3y + 6 = 0$

---

The required equation of the line is $2x - 3y + 6 = 0$.

Given:

The length of the perpendicular (normal) from the origin to the line is $p = 4$ units.

The angle that this normal makes with the positive direction of the x-axis is $\omega = 15^\circ$.

---

To Find:

The equation of the line.

---

Solution:

The equation of a line in the normal form is:

$x \cos \omega + y \sin \omega = p$

We are given $p = 4$ and $\omega = 15^\circ$. We need to find the values of $\cos 15^\circ$ and $\sin 15^\circ$.

We can express $15^\circ$ as a difference of standard angles, such as $45^\circ - 30^\circ$.

Using the angle subtraction identity for cosine:

$\cos(A - B) = \cos A \cos B + \sin A \sin B$

$\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$

$\cos 15^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right) = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

Using the angle subtraction identity for sine:

$\sin(A - B) = \sin A \cos B - \cos A \sin B$

$\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$

$\sin 15^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right) = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

Now, substitute these values and $p=4$ into the normal form equation:

$x \left(\frac{\sqrt{3} + 1}{2\sqrt{2}}\right) + y \left(\frac{\sqrt{3} - 1}{2\sqrt{2}}\right) = 4$

To simplify, multiply the entire equation by $2\sqrt{2}$:

$(\sqrt{3} + 1)x + (\sqrt{3} - 1)y = 4 \times 2\sqrt{2}$

$(\sqrt{3} + 1)x + (\sqrt{3} - 1)y = 8\sqrt{2}$

---

The required equation of the line is $(\sqrt{3} + 1)x + (\sqrt{3} - 1)y = 8\sqrt{2}$.

Given:

Fahrenheit temperature (F) and absolute temperature (K) have a linear relationship.

This relationship can be viewed as a line passing through two points:

Point 1: $(F_1, K_1) = (32, 273)$

Point 2: $(F_2, K_2) = (212, 373)$

---

To Find:

1. An equation to express K in terms of F.

2. The value of F when K = 0.

---

Solution:

1. Express K in terms of F

We can use the two-point form of a linear equation. Let's treat F as the x-coordinate and K as the y-coordinate.

The two-point form is: $\frac{K - K_1}{K_2 - K_1} = \frac{F - F_1}{F_2 - F_1}$

Substitute the given points into the formula:

$\frac{K - 273}{373 - 273} = \frac{F - 32}{212 - 32}$

$\frac{K - 273}{100} = \frac{F - 32}{180}$

To express K in terms of F, we solve for K:

$K - 273 = \frac{100}{180} (F - 32)$

Simplify the fraction $\frac{100}{180} = \frac{10}{18} = \frac{5}{9}$.

$K - 273 = \frac{5}{9} (F - 32)$

Now, add 273 to both sides:

$K = \frac{5}{9}(F - 32) + 273$

This is the required relationship between K and F.

---

2. Find the value of F when K = 0

Substitute $K = 0$ into the equation we just found:

$0 = \frac{5}{9}(F - 32) + 273$

First, subtract 273 from both sides:

$-273 = \frac{5}{9}(F - 32)$

Next, multiply both sides by $\frac{9}{5}$ to isolate the term $(F - 32)$:

$-273 \times \frac{9}{5} = F - 32$

$-\frac{2457}{5} = F - 32$

$-491.4 = F - 32$

Finally, add 32 to both sides to solve for F:

$F = -491.4 + 32$

$F = -459.4$

---

The value of F when K = 0 is -459.4.

To Find:

The equations for the x-axis and the y-axis.

---

Solution:

---

Equation of the x-axis:

The x-axis is a horizontal line. For any point that lies on the x-axis, its perpendicular distance from the x-axis is zero. This means that the y-coordinate of every point on the x-axis is 0.

For example, points like (–3, 0), (0, 0), (5, 0) are all on the x-axis.

The equation of a horizontal line is given by $y = c$, where $c$ is the constant y-coordinate. Since $y=0$ for all points on the x-axis, the equation of the x-axis is:

$y = 0$

---

Equation of the y-axis:

The y-axis is a vertical line. For any point that lies on the y-axis, its perpendicular distance from the y-axis is zero. This means that the x-coordinate of every point on the y-axis is 0.

For example, points like (0, –2), (0, 0), (0, 4) are all on the y-axis.

The equation of a vertical line is given by $x = c$, where $c$ is the constant x-coordinate. Since $x=0$ for all points on the y-axis, the equation of the y-axis is:

$x = 0$

Given:

The line passes through the point $(x_1, y_1) = (–4, 3)$.

The slope of the line is $m = \frac{1}{2}$.

---

To Find:

The equation of the line.

---

Solution:

We use the point-slope form of the equation of a line, which is:

$y - y_1 = m(x - x_1)$

Substitute the given point $(x_1, y_1) = (–4, 3)$ and slope $m = \frac{1}{2}$ into the formula:

$y - 3 = \frac{1}{2} (x - (-4))$

$y - 3 = \frac{1}{2} (x + 4)$

To eliminate the fraction, multiply both sides of the equation by 2:

$2(y - 3) = x + 4$

$2y - 6 = x + 4$

Rearrange the terms to write the equation in the general form $Ax + By + C = 0$:

$x - 2y + 4 + 6 = 0$

$x - 2y + 10 = 0$

---

The required equation of the line is $x - 2y + 10 = 0$.

Given:

The line passes through the point $(x_1, y_1) = (0, 0)$, which is the origin.

The slope of the line is $m$.

---

To Find:

The equation of the line.

---

Solution:

Method 1: Using Point-Slope Form

The point-slope form of the equation of a line is:

$y - y_1 = m(x - x_1)$

Substitute the point $(0, 0)$ and slope $m$:

$y - 0 = m(x - 0)$

$y = mx$

---

Method 2: Using Slope-Intercept Form

The slope-intercept form of the equation of a line is:

$y = mx + c$

where $c$ is the y-intercept. Since the line passes through the origin (0, 0), the y-intercept is 0.

Substitute $c = 0$ into the equation:

$y = mx + 0$

$y = mx$

---

The required equation of any line passing through the origin with slope $m$ is $y = mx$.

Given:

The line passes through the point $(x_1, y_1) = (2, 2\sqrt{3})$.

The angle of inclination with the positive x-axis is $\theta = 75^\circ$.

---

To Find:

The equation of the line.

---

Solution:

First, we need to find the slope ($m$) of the line from its angle of inclination $\theta$.

The slope is given by the relation $m = \tan \theta$.

$m = \tan 75^\circ$

To find the value of $\tan 75^\circ$, we can write $75^\circ$ as the sum of two standard angles, $45^\circ + 30^\circ$.

Using the tangent addition formula, $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:

$m = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ}$

We know that $\tan 45^\circ = 1$ and $\tan 30^\circ = \frac{1}{\sqrt{3}}$. Substituting these values:

$m = \frac{1 + \frac{1}{\sqrt{3}}}{1 - (1)\left(\frac{1}{\sqrt{3}}\right)} = \frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}$

$m = \frac{\sqrt{3}+1}{\sqrt{3}-1}$

Now, we use the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$, with the slope $m = \frac{\sqrt{3}+1}{\sqrt{3}-1}$ and the point $(x_1, y_1) = (2, 2\sqrt{3})$.

$y - 2\sqrt{3} = \frac{\sqrt{3}+1}{\sqrt{3}-1}(x - 2)$

To clear the denominator, multiply both sides by $(\sqrt{3}-1)$:

$(\sqrt{3}-1)(y - 2\sqrt{3}) = (\sqrt{3}+1)(x - 2)$

Expand both sides of the equation:

$(\sqrt{3}-1)y - 2\sqrt{3}(\sqrt{3}-1) = (\sqrt{3}+1)x - 2(\sqrt{3}+1)$

$(\sqrt{3}-1)y - (6 - 2\sqrt{3}) = (\sqrt{3}+1)x - (2\sqrt{3} + 2)$

$(\sqrt{3}-1)y - 6 + 2\sqrt{3} = (\sqrt{3}+1)x - 2\sqrt{3} - 2$

Now, rearrange the terms to bring the variable terms to one side and the constant terms to the other:

$(\sqrt{3}+1)x - (\sqrt{3}-1)y = -6 + 2\sqrt{3} + 2\sqrt{3} + 2$

$(\sqrt{3}+1)x - (\sqrt{3}-1)y = 4\sqrt{3} - 4$

Factor out 4 from the right-hand side:

$(\sqrt{3}+1)x - (\sqrt{3}-1)y = 4(\sqrt{3}-1)$

---

The required equation of the line is $(\sqrt{3}+1)x - (\sqrt{3}-1)y = 4(\sqrt{3}-1)$.

Given:

The line intersects the x-axis at a distance of 3 units to the left of the origin. This implies that the x-intercept is –3. The point of intersection on the x-axis is therefore $(–3, 0)$.

The slope of the line is $m = –2$.

---

To Find:

The equation of the line.

---

Solution:

We have a point on the line $(x_1, y_1) = (–3, 0)$ and the slope $m = –2$.

Using the point-slope form of the equation of a line:

$y - y_1 = m(x - x_1)$

Substitute the values:

$y - 0 = -2(x - (-3))$

$y = -2(x + 3)$

$y = -2x - 6$

Rearranging to the general form $Ax + By + C = 0$:

$2x + y + 6 = 0$

---

Alternate Solution:

Using the slope-intercept form $y=mx+c$. We are given $m=-2$.

$y = -2x + c$

Since the line passes through $(-3, 0)$, we can substitute these coordinates to find $c$:

$0 = -2(-3) + c$

$0 = 6 + c$

$c = -6$

Substitute $c=-6$ back into the equation:

$y = -2x - 6$

$2x + y + 6 = 0$

---

The required equation of the line is $2x + y + 6 = 0$.

Given:

The line intersects the y-axis at a distance of 2 units above the origin, which means the y-intercept is $c = 2$.

The angle of inclination with the positive x-axis is $\theta = 30^\circ$.

---

To Find:

The equation of the line.

---

Solution:

First, we find the slope ($m$) of the line from its angle of inclination.

$m = \tan \theta = \tan 30^\circ = \frac{1}{\sqrt{3}}$

Now we have the slope $m = \frac{1}{\sqrt{3}}$ and the y-intercept $c = 2$.

We can directly use the slope-intercept form of the equation of a line:

$y = mx + c$

Substitute the values of $m$ and $c$:

$y = \frac{1}{\sqrt{3}}x + 2$

To write the equation in the general form, we can eliminate the fraction by multiplying the entire equation by $\sqrt{3}$:

$\sqrt{3}y = \sqrt{3}\left(\frac{1}{\sqrt{3}}x\right) + \sqrt{3}(2)$

$\sqrt{3}y = x + 2\sqrt{3}$

Rearranging the terms:

$x - \sqrt{3}y + 2\sqrt{3} = 0$

---

Alternate Solution:

The y-intercept is 2, which means the line passes through the point $(0, 2)$. The slope is $m = \frac{1}{\sqrt{3}}$.

Using the point-slope form $y-y_1 = m(x-x_1)$:

$y - 2 = \frac{1}{\sqrt{3}}(x - 0)$

$\sqrt{3}(y - 2) = x$

$\sqrt{3}y - 2\sqrt{3} = x$

$x - \sqrt{3}y + 2\sqrt{3} = 0$

---

The required equation of the line is $x - \sqrt{3}y + 2\sqrt{3} = 0$.

Given:

The line passes through two points: $P_1(x_1, y_1) = (–1, 1)$ and $P_2(x_2, y_2) = (2, –4)$.

---

To Find:

The equation of the line.

---

Solution:

Method 1: Using Two-Point Form

The two-point form of the equation of a line is:

$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$

Substitute the coordinates of the given points:

$\frac{y - 1}{-4 - 1} = \frac{x - (-1)}{2 - (-1)}$

$\frac{y - 1}{-5} = \frac{x + 1}{3}$

Now, cross-multiply to simplify:

$3(y - 1) = -5(x + 1)$

$3y - 3 = -5x - 5$

Rearrange the terms into the general form $Ax + By + C = 0$:

$5x + 3y - 3 + 5 = 0$

$5x + 3y + 2 = 0$

---

Method 2: Using Point-Slope Form

First, calculate the slope ($m$) of the line:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 1}{2 - (-1)} = \frac{-5}{3}$

Now use the point-slope form, $y - y_1 = m(x - x_1)$, with the slope $m = -\frac{5}{3}$ and one of the points, say $(–1, 1)$:

$y - 1 = -\frac{5}{3}(x - (-1))$

$y - 1 = -\frac{5}{3}(x + 1)$

Multiply both sides by 3:

$3(y - 1) = -5(x + 1)$

$3y - 3 = -5x - 5$

$5x + 3y + 2 = 0$

---

The required equation of the line is $5x + 3y + 2 = 0$.

Given:

The length of the perpendicular (or normal) from the origin to the line is $p = 5$ units.

The angle that this perpendicular makes with the positive x-axis is $\omega = 30^\circ$.

---

To Find:

The equation of the line.

---

Solution:

We use the normal form of the equation of a line, which is ideal for this type of information. The normal form is:

$x \cos \omega + y \sin \omega = p$

We are given $p = 5$ and $\omega = 30^\circ$.

First, find the values of $\cos 30^\circ$ and $\sin 30^\circ$:

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

$\sin 30^\circ = \frac{1}{2}$

Now, substitute these values into the normal form equation:

$x \left(\frac{\sqrt{3}}{2}\right) + y \left(\frac{1}{2}\right) = 5$

To eliminate the denominators, multiply the entire equation by 2:

$\sqrt{3}x + y = 10$

To write the equation in the general form $Ax + By + C = 0$, we bring all terms to one side:

$\sqrt{3}x + y - 10 = 0$

---

The required equation of the line is $\sqrt{3}x + y - 10 = 0$.

Given:

The vertices of a triangle ∆PQR are P(2, 1), Q(–2, 3), and R(4, 5).

---

To Find:

The equation of the median through the vertex R.

---

Solution:

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. The median through the vertex R will connect R to the midpoint of the side PQ.

Let S be the midpoint of the side PQ.

We can find the coordinates of S using the midpoint formula:

$S(x, y) = \left( \frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2} \right)$

Using the coordinates of P(2, 1) and Q(–2, 3):

$x = \frac{2 + (-2)}{2} = \frac{0}{2} = 0$

$y = \frac{1 + 3}{2} = \frac{4}{2} = 2$

Thus, the coordinates of the midpoint S are (0, 2).

Now, we need to find the equation of the line passing through the points R(4, 5) and S(0, 2). This line represents the median RS.

Using the two-point form of the equation of a line:

$\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$

Let $(x_1, y_1) = R(4, 5)$ and $(x_2, y_2) = S(0, 2)$.

$\frac{y - 5}{x - 4} = \frac{2 - 5}{0 - 4} = \frac{-3}{-4} = \frac{3}{4}$

Now, we cross-multiply to simplify the equation:

$4(y - 5) = 3(x - 4)$

$4y - 20 = 3x - 12$

Rearranging the terms to the general form $Ax + By + C = 0$:

$3x - 4y - 12 + 20 = 0$

$3x - 4y + 8 = 0$

---

The equation of the median through the vertex R is $3x - 4y + 8 = 0$.

Given:

The required line passes through the point P(–3, 5).

The line is perpendicular to another line passing through the points A(2, 5) and B(–3, 6).

---

To Find:

The equation of the required line.

---

Solution:

First, we determine the slope of the line passing through points A(2, 5) and B(–3, 6). Let's call this slope $m_1$.

Using the slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$:

$m_1 = \frac{6 - 5}{-3 - 2} = \frac{1}{-5}$

Let the slope of the required line be $m_2$.

Since the required line is perpendicular to the line AB, the product of their slopes must be –1. This is the condition for perpendicular lines:

$m_1 \times m_2 = -1$

$\left(-\frac{1}{5}\right) \times m_2 = -1$

$m_2 = (-1) \times (-5) = 5$

Now we know the slope of our required line is $m_2 = 5$, and it passes through the point P(–3, 5).

We can now use the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$.

Substitute the point $(x_1, y_1) = (–3, 5)$ and the slope $m = 5$:

$y - 5 = 5(x - (-3))$

$y - 5 = 5(x + 3)$

$y - 5 = 5x + 15$

Rearrange the terms into the general form $Ax + By + C = 0$:

$5x - y + 15 + 5 = 0$

$5x - y + 20 = 0$

---

The required equation of the line is $5x - y + 20 = 0$.

Given:

A line segment joins the points A(1, 0) and B(2, 3).

The required line is perpendicular to the line segment AB.

The required line divides the segment AB in the ratio $1:n$.

---

To Find:

The equation of the required line.

---

Solution:

Step 1: Find the slope of the required line.

First, we find the slope of the line segment AB, let's call it $m_{AB}$.

$m_{AB} = \frac{y_B - y_A}{x_B - x_A} = \frac{3 - 0}{2 - 1} = \frac{3}{1} = 3$

Let the slope of the required line be $m$. Since this line is perpendicular to AB, the product of their slopes is –1.

$m \times m_{AB} = -1$

$m \times 3 = -1 \implies m = -\frac{1}{3}$

Step 2: Find the point of intersection.

The required line passes through a point P that divides the segment AB in the ratio $1:n$. We find the coordinates of P using the section formula:

$P(x, y) = \left( \frac{1 \cdot x_B + n \cdot x_A}{1 + n}, \frac{1 \cdot y_B + n \cdot y_A}{1 + n} \right)$

Substitute the coordinates of A(1, 0) and B(2, 3):

$x = \frac{1(2) + n(1)}{1 + n} = \frac{2 + n}{1 + n}$

$y = \frac{1(3) + n(0)}{1 + n} = \frac{3}{1 + n}$

So, the line passes through the point $P\left( \frac{n + 2}{n + 1}, \frac{3}{n + 1} \right)$.

Step 3: Find the equation of the line.

Now we use the point-slope form, $y - y_1 = m(x - x_1)$, with the slope $m = -\frac{1}{3}$ and the point P.

$y - \frac{3}{n + 1} = -\frac{1}{3} \left( x - \frac{n + 2}{n + 1} \right)$

To eliminate the denominators, multiply the entire equation by $3(n + 1)$:

$3(n + 1) \left( y - \frac{3}{n + 1} \right) = 3(n + 1) \left[ -\frac{1}{3} \left( x - \frac{n + 2}{n + 1} \right) \right]$

$3(n + 1)y - 3(3) = -(n + 1)\left( x - \frac{n + 2}{n + 1} \right)$

$3(n + 1)y - 9 = -(n + 1)x + (n + 2)$

Finally, rearrange the terms into the general form $Ax + By + C = 0$:

$(n + 1)x + 3(n + 1)y - 9 - (n + 2) = 0$

$(n + 1)x + 3(n + 1)y - n - 11 = 0$

---

The required equation of the line is $(n + 1)x + 3(n + 1)y - n - 11 = 0$.

Given:

A line cuts off equal intercepts on the coordinate axes.

The line passes through the point (2, 3).

---

To Find:

The equation of the line.

---

Solution:

The equation of a line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept. We are given that the intercepts are equal, so $a = b$.

There are two possible cases for this condition.

---

Case 1: The intercepts are non-zero ($a = b \neq 0$)

If $a=b$, the intercept form becomes:

$\frac{x}{a} + \frac{y}{a} = 1$

$x + y = a$

Since the line passes through the point (2, 3), these coordinates must satisfy the equation. Substitute $x=2$ and $y=3$ to find the value of $a$:

$2 + 3 = a \implies a = 5$

The intercepts are $a=5$ and $b=5$. The equation of the line is:

$x + y = 5$ or $x + y - 5 = 0$.

---

Case 2: The intercepts are zero ($a = b = 0$)

If both intercepts are zero, the line passes through the origin (0, 0).

The line must also pass through the given point (2, 3).

We can find the equation of the line passing through (0, 0) and (2, 3). The slope $m$ is:

$m = \frac{3 - 0}{2 - 0} = \frac{3}{2}$

Using the slope-intercept form $y = mx + c$, with $c=0$ (since it passes through the origin):

$y = \frac{3}{2}x$

$2y = 3x$

The equation of the line is:

$3x - 2y = 0$.

---

Conclusion:

There are two lines that satisfy the given conditions. Typically, when the term "intercepts" is used, the non-zero case is implied. However, both solutions are mathematically correct.

The equations of the lines are $x + y - 5 = 0$ and $3x - 2y = 0$.

Given:

The line passes through the point P(2, 2).

The sum of the intercepts on the coordinate axes is 9.

---

To Find:

The equation(s) of the line.

---

Solution:

Let the equation of the line in the intercept form be:

where $a$ is the x-intercept and $b$ is the y-intercept.

From the given condition, the sum of the intercepts is 9:

$a + b = 9 \implies b = 9 - a$

Since the line passes through the point (2, 2), we can substitute $x=2$ and $y=2$ into equation (i):

$\frac{2}{a} + \frac{2}{b} = 1$

Now, substitute $b = 9 - a$ into this equation:

$\frac{2}{a} + \frac{2}{9 - a} = 1$

To solve for $a$, we find a common denominator, which is $a(9-a)$, and multiply the equation by it:

$2(9 - a) + 2a = 1 \cdot a(9 - a)$

$18 - 2a + 2a = 9a - a^2$

$18 = 9a - a^2$

Rearrange this into a standard quadratic equation form ($Ax^2 + Bx + C = 0$):

$a^2 - 9a + 18 = 0$

We can factor this quadratic equation:

$a^2 - 3a - 6a + 18 = 0$

$a(a - 3) - 6(a - 3) = 0$

$(a - 3)(a - 6) = 0$

This gives two possible values for $a$: $a = 3$ or $a = 6$.

We find the corresponding values for $b$ and the resulting line equations for each case.

---

Case 1: $a = 3$

If $a = 3$, then $b = 9 - a = 9 - 3 = 6$.

The equation is $\frac{x}{3} + \frac{y}{6} = 1$.

Multiplying by 6 to clear the fractions, we get $2x + y = 6$, or $2x + y - 6 = 0$.

---

Case 2: $a = 6$

If $a = 6$, then $b = 9 - a = 9 - 6 = 3$.

The equation is $\frac{x}{6} + \frac{y}{3} = 1$.

Multiplying by 6 to clear the fractions, we get $x + 2y = 6$, or $x + 2y - 6 = 0$.

---

There are two lines that satisfy the given conditions. Their equations are $2x + y - 6 = 0$ and $x + 2y - 6 = 0$.

This problem has two parts. We will solve for each line separately.

---

Part 1: Equation of the first line

Given:

The line passes through the point (0, 2).

The angle of inclination with the positive x-axis is $\theta = \frac{2\pi}{3}$ radians.

Solution:

First, we find the slope ($m_1$) of the line. The slope is the tangent of the angle of inclination.

$m_1 = \tan\left(\frac{2\pi}{3}\right) = \tan(120^\circ)$

$m_1 = \tan(180^\circ - 60^\circ) = -\tan(60^\circ) = -\sqrt{3}$

The line passes through the point (0, 2). This means the y-intercept is $c_1 = 2$.

Using the slope-intercept form of a line, $y = mx + c$:

$y = -\sqrt{3}x + 2$

Rearranging to the general form:

$\sqrt{3}x + y - 2 = 0$

---

Part 2: Equation of the second (parallel) line

Given:

The second line is parallel to the first line.

It crosses the y-axis at a distance of 2 units below the origin.

Solution:

Since the second line is parallel to the first, it must have the same slope. So, the slope of the second line is $m_2 = m_1 = -\sqrt{3}$.

The line crosses the y-axis 2 units below the origin, which means its y-intercept is $c_2 = -2$.

Again, using the slope-intercept form, $y = mx + c$:

$y = -\sqrt{3}x + (-2)$

$y = -\sqrt{3}x - 2$

Rearranging to the general form:

$\sqrt{3}x + y + 2 = 0$

---

Final Answer:

The equation of the first line is $\sqrt{3}x + y - 2 = 0$.

The equation of the parallel line is $\sqrt{3}x + y + 2 = 0$.

Given:

A line has a perpendicular drawn to it from the origin O(0, 0).

This perpendicular meets the line at the point P(–2, 9).

---

To Find:

The equation of the line.

---

Solution:

The situation describes a line that passes through the point P(–2, 9) and is perpendicular to the line segment OP, where O is the origin (0, 0).

Step 1: Find the slope of the perpendicular line segment OP.

Let the slope of the line segment OP be $m_{OP}$. Using the slope formula with points O(0, 0) and P(–2, 9):

$m_{OP} = \frac{9 - 0}{-2 - 0} = -\frac{9}{2}$

Step 2: Find the slope of the required line.

Let the slope of the required line be $m$. Since the line is perpendicular to the segment OP, the product of their slopes is –1.

$m \times m_{OP} = -1$

$m \times \left(-\frac{9}{2}\right) = -1$

$m = \frac{-1}{-9/2} = \frac{2}{9}$

Step 3: Find the equation of the line.

We now know that the required line passes through the point P(–2, 9) and has a slope of $m = \frac{2}{9}$.

Using the point-slope form of a line, $y - y_1 = m(x - x_1)$:

$y - 9 = \frac{2}{9}(x - (-2))$

$y - 9 = \frac{2}{9}(x + 2)$

To eliminate the fraction, multiply both sides by 9:

$9(y - 9) = 2(x + 2)$

$9y - 81 = 2x + 4$

Rearranging the terms into the general form $Ax + By + C = 0$:

$2x - 9y + 4 + 81 = 0$

$2x - 9y + 85 = 0$

---

The required equation of the line is $2x - 9y + 85 = 0$.

Given:

The length L of a copper rod is a linear function of its Celsius temperature C.

This linear relationship passes through two points, treating (C, L) as coordinates:

Point 1: $(C_1, L_1) = (20, 124.942)$

Point 2: $(C_2, L_2) = (110, 125.134)$

---

To Find:

An equation that expresses L in terms of C.

---

Solution:

Since the relationship is linear, we can use the two-point form of a linear equation to establish the relationship between L and C.

The two-point form is given by:

$\frac{L - L_1}{C - C_1} = \frac{L_2 - L_1}{C_2 - C_1}$

Substitute the coordinates of the two given points:

$\frac{L - 124.942}{C - 20} = \frac{125.134 - 124.942}{110 - 20}$

Now, calculate the value of the right-hand side, which represents the slope ($m$) of the line:

$\frac{L - 124.942}{C - 20} = \frac{0.192}{90}$

To express L in terms of C, we can rearrange the equation into the point-slope form, $L - L_1 = m(C - C_1)$:

$L - 124.942 = \frac{0.192}{90}(C - 20)$

Solving for L, we get:

$L = \frac{0.192}{90}(C - 20) + 124.942$

This is a complete and correct expression for L in terms of C.

Note: The fraction $\frac{0.192}{90}$ is mathematically correct. The expression $\frac{192}{90}$ would be incorrect as it is 1000 times larger.

---

Alternate Simplified Form:

We can further simplify the equation into the standard slope-intercept form, $L = mC + b$.

First, simplify the slope $m = \frac{0.192}{90}$:

$m = \frac{192}{90000} = \frac{4}{1875}$

Now, substitute this simplified slope back into the point-slope equation:

$L = \frac{4}{1875}(C - 20) + 124.942$

Expand the equation:

$L = \frac{4}{1875}C - \frac{80}{1875} + 124.942$

$L = \frac{4}{1875}C + \left(124.942 - \frac{80}{1875}\right)$

To calculate the constant term (the L-intercept, b):

$b = \frac{124942}{1000} - \frac{80}{1875} = \frac{62471}{500} - \frac{16}{375} = \frac{187413 - 64}{1500} = \frac{187349}{1500}$

So, the fully simplified equation is:

$L = \frac{4}{1875}C + \frac{187349}{1500}$

---

The relationship expressed in a form that directly shows the experimental data is:

$L = \frac{0.192}{90}(C - 20) + 124.942$

Given:

Let the selling price be $P$ (in $\textsf{₹}$/litre) and the weekly demand be $D$ (in litres).

The relationship between $P$ and $D$ is linear.

We are given two points $(P, D)$ that lie on this line:

Point 1: $(P_1, D_1) = (14, 980)$

Point 2: $(P_2, D_2) = (16, 1220)$

---

To Find:

The demand ($D$) when the price ($P$) is $\textsf{₹}$ 17/litre.

---

Solution:

Step 1: Find the equation of the line relating Demand (D) and Price (P).

First, we calculate the slope ($m$) of the line. The slope represents the change in demand per unit change in price.

$m = \frac{D_2 - D_1}{P_2 - P_1} = \frac{1220 - 980}{16 - 14} = \frac{240}{2} = 120$

Now, we use the point-slope form, $D - D_1 = m(P - P_1)$, with the slope $m=120$ and Point 1 $(14, 980)$:

$D - 980 = 120(P - 14)$

We can simplify this equation to express $D$ in terms of $P$:

$D - 980 = 120P - 1680$

$D = 120P - 1680 + 980$

$D = 120P - 700$

This is the linear relationship between demand and price.

Step 2: Calculate the demand for a price of $\textsf{₹}$ 17/litre.

We need to find the value of $D$ when $P = 17$. Substitute $P=17$ into the equation:

$D = 120(17) - 700$

$D = 2040 - 700$

$D = 1340$

---

Therefore, the owner could sell 1340 litres of milk weekly at a price of $\textsf{₹}$ 17/litre.

Given:

A line segment is intercepted between the x and y axes.

The midpoint of this line segment is the point P(a, b).

---

To Prove:

The equation of the line is $\frac{x}{a} + \frac{y}{b} = 2$.

---

Proof:

Let the line intersect the x-axis at point A and the y-axis at point B.

Since point A lies on the x-axis, its y-coordinate is 0. Let its coordinates be A(p, 0). Here, 'p' is the x-intercept.

Since point B lies on the y-axis, its x-coordinate is 0. Let its coordinates be B(0, q). Here, 'q' is the y-intercept.

We are given that P(a, b) is the midpoint of the segment AB. Using the midpoint formula:

Coordinates of midpoint = $\left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right)$

$P(a, b) = \left( \frac{p + 0}{2}, \frac{0 + q}{2} \right) = \left( \frac{p}{2}, \frac{q}{2} \right)$

By comparing the coordinates, we get:

$a = \frac{p}{2} \implies p = 2a$

$b = \frac{q}{2} \implies q = 2b$

This means the x-intercept of the line is $2a$ and the y-intercept is $2b$.

Now, we write the equation of the line using the intercept form:

$\frac{x}{\text{x-intercept}} + \frac{y}{\text{y-intercept}} = 1$

Substitute the intercepts we found:

$\frac{x}{2a} + \frac{y}{2b} = 1$

To get the equation in the required form, we multiply both sides by 2:

$2 \left( \frac{x}{2a} + \frac{y}{2b} \right) = 2 \times 1$

$\frac{x}{a} + \frac{y}{b} = 2$

Hence, Proved.

Given:

A line segment is intercepted between the coordinate axes.

The point R(h, k) divides this line segment in the ratio 1:2.

---

To Find:

The equation of the line.

---

Solution:

Let the line intersect the x-axis at point A and the y-axis at point B.

Let the coordinates of A be (p, 0) and the coordinates of B be (0, q). Here, 'p' is the x-intercept and 'q' is the y-intercept.

The point R(h, k) divides the line segment AB in the ratio 1:2. We assume the division is such that the segment closer to the x-axis (AR) is the smaller part, i.e., AR:RB = 1:2.

Using the section formula:

$R(x, y) = \left( \frac{m x_B + n x_A}{m + n}, \frac{m y_B + n y_A}{m + n} \right)$

Here, $(x_A, y_A) = (p, 0)$, $(x_B, y_B) = (0, q)$, and the ratio is $m:n = 1:2$.

Substitute these values to find h and k:

$h = \frac{1(0) + 2(p)}{1 + 2} = \frac{2p}{3}$

$k = \frac{1(q) + 2(0)}{1 + 2} = \frac{q}{3}$

Now, we express the intercepts, p and q, in terms of h and k:

From $h = \frac{2p}{3}$, we get $p = \frac{3h}{2}$. (This is the x-intercept)

From $k = \frac{q}{3}$, we get $q = 3k$. (This is the y-intercept)

Now, substitute these intercepts into the intercept form of the equation of a line:

$\frac{x}{\text{x-intercept}} + \frac{y}{\text{y-intercept}} = 1$

$\frac{x}{3h/2} + \frac{y}{3k} = 1$

Simplify the first term:

$\frac{2x}{3h} + \frac{y}{3k} = 1$

To write the equation in a simpler form, we can find a common denominator, which is $3hk$, and multiply the entire equation by it:

$3hk \left( \frac{2x}{3h} + \frac{y}{3k} \right) = 3hk \times 1$

$k(2x) + h(y) = 3hk$

$2kx + hy = 3hk$

---

The required equation of the line is $2kx + hy = 3hk$.

Given:

Three points: A(3, 0), B(–2, –2), and C(8, 2).

---

To Prove:

The points A, B, and C are collinear.

---

Proof:

The strategy is to find the equation of the line that passes through any two of the given points, and then show that the third point also lies on this line (i.e., its coordinates satisfy the equation).

Step 1: Find the equation of the line passing through points A(3, 0) and B(–2, –2).

First, we find the slope ($m$) of the line AB:

$m = \frac{y_B - y_A}{x_B - x_A} = \frac{-2 - 0}{-2 - 3} = \frac{-2}{-5} = \frac{2}{5}$

Now, using the point-slope form $y - y_1 = m(x - x_1)$ with point A(3, 0):

$y - 0 = \frac{2}{5}(x - 3)$

$5y = 2(x - 3)$

$5y = 2x - 6$

The equation of the line through A and B is $2x - 5y - 6 = 0$.

Step 2: Check if point C(8, 2) satisfies this equation.

Substitute $x = 8$ and $y = 2$ into the equation $2x - 5y - 6 = 0$:

LHS = $2(8) - 5(2) - 6$

LHS = $16 - 10 - 6$

LHS = $6 - 6 = 0$

Since the result is 0, which is equal to the RHS of the equation, the point C(8, 2) lies on the line passing through A and B.

Since all three points lie on the same line, they are collinear.

Hence, Proved.

---

Alternate Solution (Using Slopes):

Three points A, B, and C are collinear if the slope of AB is equal to the slope of BC.

Slope of AB: $m_{AB} = \frac{-2 - 0}{-2 - 3} = \frac{-2}{-5} = \frac{2}{5}$

Slope of BC: $m_{BC} = \frac{2 - (-2)}{8 - (-2)} = \frac{2 + 2}{8 + 2} = \frac{4}{10} = \frac{2}{5}$

Since $m_{AB} = m_{BC}$ and they share a common point B, the points A, B, and C must lie on the same straight line.

Therefore, the points are collinear.

Given:

The equation of a line is $3x – 4y + 10 = 0$.

---

To Find:

(i) The slope of the line.

(ii) The x-intercept and y-intercept of the line.

---

Solution:

---

(i) Finding the Slope

To find the slope, we can convert the given equation into the slope-intercept form, which is $y = mx + c$, where 'm' is the slope.

$3x – 4y + 10 = 0$

Isolate the y-term:

$4y = 3x + 10$

Divide the entire equation by 4:

$y = \frac{3}{4}x + \frac{10}{4}$

$y = \frac{3}{4}x + \frac{5}{2}$

By comparing this with $y = mx + c$, we can see that the slope is:

Slope (m) = $\frac{3}{4}$

---

(ii) Finding the x and y-intercepts

To find the y-intercept:

The y-intercept is the point where the line crosses the y-axis, which means the x-coordinate is 0. We can set $x=0$ in the original equation.

$3(0) – 4y + 10 = 0$

$–4y = –10$

$y = \frac{-10}{-4} = \frac{5}{2}$

Alternatively, from the slope-intercept form $y = \frac{3}{4}x + \frac{5}{2}$, the y-intercept 'c' is directly visible.

y-intercept = $\frac{5}{2}$

To find the x-intercept:

The x-intercept is the point where the line crosses the x-axis, which means the y-coordinate is 0. We can set $y=0$ in the original equation.

$3x – 4(0) + 10 = 0$

$3x + 10 = 0$

$3x = –10$

$x = -\frac{10}{3}$

x-intercept = $-\frac{10}{3}$

---

Alternate Method for Intercepts:

Convert the equation $3x - 4y + 10 = 0$ to the intercept form $\frac{x}{a} + \frac{y}{b} = 1$, where 'a' is the x-intercept and 'b' is the y-intercept.

$3x - 4y = -10$

Divide the equation by -10 to make the right side equal to 1:

$\frac{3x}{-10} - \frac{4y}{-10} = 1$

$\frac{x}{-10/3} + \frac{y}{10/4} = 1$

$\frac{x}{-10/3} + \frac{y}{5/2} = 1$

Comparing this with $\frac{x}{a} + \frac{y}{b} = 1$, we get:

x-intercept (a) = $-\frac{10}{3}$

y-intercept (b) = $\frac{5}{2}$

Given:

The equation of the line is $\sqrt{3}x + y - 8 = 0$.

---

To Find:

1. The normal form of the equation.

2. The value of $p$ (the perpendicular distance from the origin).

3. The value of $\omega$ (the angle the normal makes with the positive x-axis).

---

Solution:

The normal form of a line is given by $x \cos \omega + y \sin \omega = p$, where $p > 0$.

The given equation is $\sqrt{3}x + y - 8 = 0$.

Step 1: Rewrite the equation in the form $Ax + By = C$, with $C>0$.

$\sqrt{3}x + y = 8$

Here, $A = \sqrt{3}$, $B = 1$, and the constant term on the right side is positive, as required.

Step 2: Find the value of $\sqrt{A^2 + B^2}$.

$\sqrt{A^2 + B^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$

Step 3: Divide the entire equation by $\sqrt{A^2 + B^2}$.

Dividing the equation $\sqrt{3}x + y = 8$ by 2, we get:

$\frac{\sqrt{3}}{2}x + \frac{1}{2}y = \frac{8}{2}$

$\frac{\sqrt{3}}{2}x + \frac{1}{2}y = 4$

This is the required normal form of the equation.

Step 4: Compare with $x \cos \omega + y \sin \omega = p$ to find $p$ and $\omega$.

By comparing the two equations, we can identify the values:

$\cos \omega = \frac{\sqrt{3}}{2}$

$\sin \omega = \frac{1}{2}$

$p = 4$

Since both $\cos \omega$ and $\sin \omega$ are positive, the angle $\omega$ must be in the first quadrant.

The angle for which $\cos \omega = \frac{\sqrt{3}}{2}$ and $\sin \omega = \frac{1}{2}$ is $\omega = 30^\circ$ or $\frac{\pi}{6}$ radians.

---

Final Answer:

The normal form of the equation is $\frac{\sqrt{3}}{2}x + \frac{1}{2}y = 4$.

The value of $p$ is 4.

The value of $\omega$ is $30^\circ$ (or $\frac{\pi}{6}$).

Given:

The equations of two lines are:

Line 1 ($L_1$): $y - \sqrt{3}x - 5 = 0$

Line 2 ($L_2$): $\sqrt{3}y - x + 6 = 0$

---

To Find:

The angle between the two lines.

---

Solution:

The angle between two lines can be found using their slopes. Let the acute angle between the lines be $\theta$. The formula relating the angle to the slopes ($m_1$ and $m_2$) is:

$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

Step 1: Find the slope of each line.

For Line 1: $y - \sqrt{3}x - 5 = 0$

Rearrange into slope-intercept form ($y = mx + c$):

$y = \sqrt{3}x + 5$

The slope of the first line is $m_1 = \sqrt{3}$.

For Line 2: $\sqrt{3}y - x + 6 = 0$

Rearrange into slope-intercept form:

$\sqrt{3}y = x - 6$

$y = \frac{1}{\sqrt{3}}x - \frac{6}{\sqrt{3}}$

The slope of the second line is $m_2 = \frac{1}{\sqrt{3}}$.

Step 2: Apply the angle formula.

$\tan \theta = \left| \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + (\sqrt{3})\left(\frac{1}{\sqrt{3}}\right)} \right|$

Simplify the numerator:

$\sqrt{3} - \frac{1}{\sqrt{3}} = \frac{(\sqrt{3})^2 - 1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$

Simplify the denominator:

$1 + (\sqrt{3})\left(\frac{1}{\sqrt{3}}\right) = 1 + 1 = 2$

Substitute these back into the tangent formula:

$\tan \theta = \left| \frac{2/\sqrt{3}}{2} \right| = \left| \frac{1}{\sqrt{3}} \right| = \frac{1}{\sqrt{3}}$

Step 3: Find the angle $\theta$.

The angle whose tangent is $\frac{1}{\sqrt{3}}$ is $30^\circ$.

$\theta = 30^\circ$ or $\frac{\pi}{6}$ radians.

---

The acute angle between the given lines is $30^\circ$.

Given:

The equations of two lines are:

It is given that $b_1 \neq 0$ and $b_2 \neq 0$.

---

To Prove:

(i) The lines are parallel if $\frac{a_1}{b_1} = \frac{a_2}{b_2}$.

(ii) The lines are perpendicular if $a_1 a_2 + b_1 b_2 = 0$.

---

Proof:

First, we find the slopes of the two lines. Since $b_1$ and $b_2$ are non-zero, we can rearrange the equations into the slope-intercept form ($y = mx + c$).

Slope of Line 1 ($m_1$):

$a_1x + b_1y + c_1 = 0 \implies b_1y = -a_1x - c_1 \implies y = -\frac{a_1}{b_1}x - \frac{c_1}{b_1}$

So, $m_1 = -\frac{a_1}{b_1}$.

Slope of Line 2 ($m_2$):

$a_2x + b_2y + c_2 = 0 \implies b_2y = -a_2x - c_2 \implies y = -\frac{a_2}{b_2}x - \frac{c_2}{b_2}$

So, $m_2 = -\frac{a_2}{b_2}$.

---

(i) Condition for Parallel Lines

Two non-vertical lines are parallel if and only if their slopes are equal ($m_1 = m_2$).

$-\frac{a_1}{b_1} = -\frac{a_2}{b_2}$

Multiplying both sides by –1, we get:

$\frac{a_1}{b_1} = \frac{a_2}{b_2}$

(Note: This is equivalent to the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ provided $a_2, b_2 \neq 0$)

Thus, the two lines are parallel if $\frac{a_1}{b_1} = \frac{a_2}{b_2}$.

Hence, Proved.

---

(ii) Condition for Perpendicular Lines

Two non-vertical lines are perpendicular if and only if the product of their slopes is –1 ($m_1 m_2 = -1$).

$\left(-\frac{a_1}{b_1}\right) \times \left(-\frac{a_2}{b_2}\right) = -1$

$\frac{a_1 a_2}{b_1 b_2} = -1$

Multiply both sides by $b_1 b_2$ (which is non-zero):

$a_1 a_2 = -b_1 b_2$

Adding $b_1 b_2$ to both sides gives the required condition:

$a_1 a_2 + b_1 b_2 = 0$

Thus, the two lines are perpendicular if $a_1 a_2 + b_1 b_2 = 0$.

Hence, Proved.

Given:

The equation of a given line is $L_1: x - 2y + 3 = 0$.

The required line, $L_2$, passes through the point P(1, –2).

The line $L_2$ is perpendicular to the line $L_1$.

---

To Find:

The equation of the line $L_2$.

---

Solution:

Step 1: Find the slope of the given line ($m_1$).

We rearrange the equation $x - 2y + 3 = 0$ into the slope-intercept form ($y = mx + c$).

$2y = x + 3$

$y = \frac{1}{2}x + \frac{3}{2}$

The slope of the given line is $m_1 = \frac{1}{2}$.

Step 2: Find the slope of the required line ($m_2$).

Since the required line is perpendicular to the given line, the product of their slopes is –1.

$m_1 \times m_2 = -1$

$\frac{1}{2} \times m_2 = -1 \implies m_2 = -2$

Step 3: Find the equation of the required line.

The required line passes through the point $(x_1, y_1) = (1, –2)$ and has a slope $m_2 = -2$.

Using the point-slope form, $y - y_1 = m_2(x - x_1)$:

$y - (–2) = -2(x - 1)$

$y + 2 = -2x + 2$

Rearranging the terms to the general form $Ax + By + C = 0$:

$2x + y + 2 - 2 = 0$

$2x + y = 0$

---

Alternate Solution:

The equation of any line perpendicular to $Ax + By + C = 0$ can be written in the form $Bx - Ay + k = 0$ for some constant $k$.

The given line is $x - 2y + 3 = 0$. Here, $A=1$ and $B=-2$.

The equation of a line perpendicular to it is $-2x - 1y + k = 0$, which can be simplified to $2x + y - k = 0$. Let $k' = -k$.

So, the equation is $2x + y + k' = 0$.

This line must pass through the point (1, –2). Substitute these coordinates to find $k'$:

$2(1) + (–2) + k' = 0$

$2 - 2 + k' = 0 \implies k' = 0$

Substituting $k' = 0$ back into the equation gives:

$2x + y + 0 = 0$

$2x + y = 0$

---

The required equation of the line is $2x + y = 0$.

Given:

The point is $P(x_1, y_1) = (3, –5)$.

The equation of the line is $L: 3x – 4y – 26 = 0$.

---

To Find:

The perpendicular distance from point P to line L.

---

Solution:

The formula for the perpendicular distance ($d$) from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is:

$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

From the given line equation, $3x – 4y – 26 = 0$, we identify the coefficients:

$A = 3$, $B = –4$, $C = –26$

The coordinates of the given point are:

$x_1 = 3$, $y_1 = –5$

Now, substitute these values into the distance formula:

$d = \frac{|(3)(3) + (–4)(–5) + (–26)|}{\sqrt{3^2 + (–4)^2}}$

First, calculate the value inside the absolute value in the numerator:

$|9 + 20 – 26| = |29 – 26| = |3| = 3$

Next, calculate the value of the denominator:

$\sqrt{3^2 + (–4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Finally, compute the distance:

$d = \frac{3}{5}$

---

The distance of the point (3, –5) from the line 3x – 4y – 26 = 0 is $\frac{3}{5}$ units.

Given:

The equations of two parallel lines are:

$L_1: 3x – 4y + 7 = 0$

$L_2: 3x – 4y + 5 = 0$

---

To Find:

The perpendicular distance between these two lines.

---

Solution:

The distance ($d$) between two parallel lines written in the form $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula:

$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$

First, we identify the coefficients from the given equations. Both lines have the same coefficients for x and y, confirming they are parallel.

$A = 3$

$B = –4$

For the first line, $C_1 = 7$.

For the second line, $C_2 = 5$.

Now, substitute these values into the distance formula:

$d = \frac{|7 - 5|}{\sqrt{3^2 + (–4)^2}}$

Calculate the numerator:

$|7 - 5| = |2| = 2$

Calculate the denominator:

$\sqrt{3^2 + (–4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Compute the distance:

$d = \frac{2}{5}$

---

The distance between the two parallel lines is $\frac{2}{5}$ units.

The slope-intercept form of a linear equation is $y = mx + c$, where 'm' represents the slope and 'c' represents the y-intercept.

---

(i) x + 7y = 0

To convert the given equation to slope-intercept form, we need to solve for $y$.

$7y = -x$

$y = -\frac{1}{7}x$

To match the form $y = mx + c$, we can write this as:

$y = \left(-\frac{1}{7}\right)x + 0$

By comparing, we find:

Slope-intercept form: $y = -\frac{1}{7}x$

Slope (m): $-\frac{1}{7}$

Y-intercept (c): $0$

---

(ii) 6x + 3y – 5 = 0

Solving for $y$:

$3y = -6x + 5$

$y = \frac{-6x + 5}{3}$

$y = -2x + \frac{5}{3}$

By comparing with $y = mx + c$, we find:

Slope-intercept form: $y = -2x + \frac{5}{3}$

Slope (m): $-2$

Y-intercept (c): $\frac{5}{3}$

---

(iii) y = 0

This equation is already in a form that can be compared to $y=mx+c$. It represents the x-axis.

We can write it as:

$y = (0)x + 0$

By comparing, we find:

Slope-intercept form: $y = 0$

Slope (m): $0$

Y-intercept (c): $0$

The intercept form of a linear equation is $\frac{x}{a} + \frac{y}{b} = 1$, where 'a' is the x-intercept and 'b' is the y-intercept.

---

(i) 3x + 2y – 12 = 0

To convert to intercept form, first move the constant term to the right side.

$3x + 2y = 12$

Next, divide the entire equation by the constant term (12) to make the right side equal to 1.

$\frac{3x}{12} + \frac{2y}{12} = \frac{12}{12}$

Simplify the fractions to get the standard form:

$\frac{x}{4} + \frac{y}{6} = 1$

By comparing with $\frac{x}{a} + \frac{y}{b} = 1$, we find:

Intercept form: $\frac{x}{4} + \frac{y}{6} = 1$

x-intercept (a): $4$

y-intercept (b): $6$

---

(ii) 4x – 3y = 6

The constant is already on the right side. Divide the entire equation by 6.

$\frac{4x}{6} - \frac{3y}{6} = \frac{6}{6}$

Simplify and rearrange to match the standard form:

$\frac{2x}{3} - \frac{y}{2} = 1$

$\frac{x}{3/2} + \frac{y}{-2} = 1$

By comparing with $\frac{x}{a} + \frac{y}{b} = 1$, we find:

Intercept form: $\frac{x}{3/2} + \frac{y}{-2} = 1$

x-intercept (a): $\frac{3}{2}$

y-intercept (b): $-2$

---

(iii) 3y + 2 = 0

First, solve for y:

$3y = -2 \implies y = -\frac{2}{3}$

This equation represents a horizontal line.

y-intercept: The line crosses the y-axis at $y = -2/3$. So, the y-intercept is $-\frac{2}{3}$.

x-intercept: Since the line is parallel to the x-axis (its slope is 0), it never crosses the x-axis. Therefore, there is no x-intercept.

Because there is no finite x-intercept, the equation cannot be reduced to the standard intercept form $\frac{x}{a} + \frac{y}{b} = 1$.

The normal form of a line is $x \cos \omega + y \sin \omega = p$, where $p$ is the perpendicular distance from the origin ($p>0$) and $\omega$ is the angle the normal makes with the positive x-axis.

---

(i) x – $\sqrt{3}$y + 8 = 0

Rearrange to the form $Ax+By=C$: $x - \sqrt{3}y = -8$.

Since $p$ must be positive, the constant on the right side must be positive. Multiply the equation by -1:

$-x + \sqrt{3}y = 8$. Here $A=-1, B=\sqrt{3}$.

Divide by $\sqrt{A^2 + B^2} = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2$.

$-\frac{1}{2}x + \frac{\sqrt{3}}{2}y = 4$

Comparing with $x \cos \omega + y \sin \omega = p$:

$\cos \omega = -1/2$, $\sin \omega = \sqrt{3}/2$, and $p=4$.

Since cosine is negative and sine is positive, $\omega$ is in the second quadrant. The angle is $\omega=120^\circ$.

Normal form: $-\frac{1}{2}x + \frac{\sqrt{3}}{2}y = 4$

Perpendicular distance (p): 4

Angle ($\omega$): $120^\circ$

---

(ii) y – 2 = 0

Rearrange to the form $Ax+By=C$: $0x + 1y = 2$. Here $A=0, B=1$.

The constant on the right (2) is already positive.

Divide by $\sqrt{A^2 + B^2} = \sqrt{0^2 + 1^2} = 1$. The equation remains the same.

$0x + 1y = 2$

Comparing with $x \cos \omega + y \sin \omega = p$:

$\cos \omega = 0$, $\sin \omega = 1$, and $p=2$.

The angle for which cosine is 0 and sine is 1 is $\omega=90^\circ$.

Normal form: $y = 2$

Perpendicular distance (p): 2

Angle ($\omega$): $90^\circ$

---

(iii) x – y = 4

The equation is in the form $Ax+By=C$ with $C>0$: $1x - 1y = 4$. Here $A=1, B=-1$.

Divide by $\sqrt{A^2 + B^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.

$\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y = \frac{4}{\sqrt{2}}$

Rationalizing the constant term: $\frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.

The normal form is $\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y = 2\sqrt{2}$.

Comparing with $x \cos \omega + y \sin \omega = p$:

$\cos \omega = 1/\sqrt{2}$, $\sin \omega = -1/\sqrt{2}$, and $p=2\sqrt{2}$.

Since cosine is positive and sine is negative, $\omega$ is in the fourth quadrant. The angle is $\omega=315^\circ$.

Normal form: $\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y = 2\sqrt{2}$

Perpendicular distance (p): $2\sqrt{2}$

Angle ($\omega$): $315^\circ$

Given:

The point is P(–1, 1).

The equation of the line is $12(x + 6) = 5(y – 2)$.

---

To Find:

The perpendicular distance from the point P to the given line.

---

Solution:

Step 1: Convert the line equation to the general form $Ax + By + C = 0$.

$12(x + 6) = 5(y – 2)$

$12x + 72 = 5y – 10$

$12x – 5y + 72 + 10 = 0$

$12x – 5y + 82 = 0$

Step 2: Use the distance formula.

The distance ($d$) from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by:

$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Here, $A = 12$, $B = –5$, $C = 82$, and the point is $(x_1, y_1) = (–1, 1)$.

Substitute these values into the formula:

$d = \frac{|12(–1) + (–5)(1) + 82|}{\sqrt{12^2 + (–5)^2}}$

$d = \frac{|–12 – 5 + 82|}{\sqrt{144 + 25}}$

$d = \frac{|65|}{\sqrt{169}}$

$d = \frac{65}{13}$

$d = 5$

---

The distance of the point (–1, 1) from the line is 5 units.

Given:

The equation of the line is $\frac{x}{3} + \frac{y}{4} = 1$.

The points to be found lie on the x-axis.

The distance from these points to the line is 4 units.

---

To Find:

The coordinates of the points on the x-axis.

---

Solution:

Step 1: Convert the line equation to the general form $Ax + By + C = 0$.

Multiply the equation $\frac{x}{3} + \frac{y}{4} = 1$ by the LCM of the denominators (12):

$4x + 3y = 12 \implies 4x + 3y - 12 = 0$

Step 2: Define a general point on the x-axis.

Any point on the x-axis has a y-coordinate of 0. Let the point be P(a, 0).

Step 3: Use the distance formula and solve for the unknown coordinate.

The distance ($d$) from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.

Here, $A = 4, B = 3, C = -12$, the point is $(a, 0)$, and the distance is $d = 4$.

$4 = \frac{|4(a) + 3(0) - 12|}{\sqrt{4^2 + 3^2}}$

$4 = \frac{|4a - 12|}{\sqrt{16 + 9}} = \frac{|4a - 12|}{\sqrt{25}} = \frac{|4a - 12|}{5}$

Multiply both sides by 5:

$20 = |4a - 12|$

This absolute value equation leads to two possible cases:

Case 1: $4a - 12 = 20$

$4a = 32 \implies a = 8$

Case 2: $4a - 12 = -20$

$4a = -8 \implies a = -2$

The two possible values for the x-coordinate are 8 and -2.

---

The required points on the x-axis are (8, 0) and (–2, 0).

The distance ($d$) between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula:

$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$

---

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

The lines are in the required form. We can identify the coefficients:

$A = 15$, $B = 8$, $C_1 = –34$, $C_2 = 31$.

Substitute these values into the formula:

$d = \frac{|–34 - 31|}{\sqrt{15^2 + 8^2}}$

$d = \frac{|-65|}{\sqrt{225 + 64}} = \frac{65}{\sqrt{289}} = \frac{65}{17}$

The distance between the lines is $\frac{65}{17}$ units.

---

(ii) l(x + y) + p = 0 and l(x + y) – r = 0

First, expand the equations to the form $Ax + By + C = 0$.

$L_1: lx + ly + p = 0$

$L_2: lx + ly – r = 0$

We can identify the coefficients:

$A = l$, $B = l$, $C_1 = p$, $C_2 = –r$.

Substitute these values into the formula:

$d = \frac{|p - (–r)|}{\sqrt{l^2 + l^2}}$

$d = \frac{|p + r|}{\sqrt{2l^2}} = \frac{|p + r|}{\sqrt{2}|l|}$

The distance between the lines is $\frac{|p + r|}{\sqrt{2}|l|}$ units (assuming $l \neq 0$).

Given:

A line with equation $3x - 4y + 2 = 0$.

A point P(–2, 3).

---

To Find:

The equation of a new line that is parallel to the given line and passes through the given point.

---

Solution:

Method 1: Using Slopes

First, find the slope of the given line by converting it to slope-intercept form ($y=mx+c$).

$3x - 4y + 2 = 0 \implies 4y = 3x + 2 \implies y = \frac{3}{4}x + \frac{1}{2}$

The slope of the given line is $m = \frac{3}{4}$.

Since the required line is parallel, it must have the same slope, $m = \frac{3}{4}$.

Now use the point-slope form, $y - y_1 = m(x - x_1)$, with the point (–2, 3) and slope $\frac{3}{4}$.

$y - 3 = \frac{3}{4}(x - (–2))$

$4(y - 3) = 3(x + 2)$

$4y - 12 = 3x + 6$

$3x - 4y + 18 = 0$

---

Method 2: Using the Form of Parallel Lines

Any line parallel to $Ax + By + C = 0$ has the form $Ax + By + k = 0$ for some constant $k$.

The given line is $3x - 4y + 2 = 0$.

Therefore, the equation of the parallel line is $3x - 4y + k = 0$.

Since this line passes through the point (–2, 3), we can substitute these coordinates to find $k$.

$3(–2) - 4(3) + k = 0$

$-6 - 12 + k = 0 \implies -18 + k = 0 \implies k = 18$

Substitute $k=18$ back into the equation.

$3x - 4y + 18 = 0$

---

The equation of the required line is $3x - 4y + 18 = 0$.

Given:

A line with equation $x – 7y + 5 = 0$.

The required line is perpendicular to this line.

The required line has an x-intercept of 3.

---

To Find:

The equation of the required line.

---

Solution:

Method 1: Using Slopes

First, find the slope of the given line: $x – 7y + 5 = 0 \implies 7y = x + 5 \implies y = \frac{1}{7}x + \frac{5}{7}$.

The slope of the given line is $m_1 = \frac{1}{7}$.

The slope of the required perpendicular line, $m_2$, is the negative reciprocal of $m_1$.

$m_2 = -\frac{1}{m_1} = -\frac{1}{1/7} = -7$.

The required line has an x-intercept of 3, which means it passes through the point (3, 0).

Using the point-slope form, $y - y_1 = m_2(x - x_1)$:

$y - 0 = -7(x - 3)$

$y = -7x + 21$

$7x + y - 21 = 0$

---

Method 2: Using the Form of Perpendicular Lines

Any line perpendicular to $Ax + By + C = 0$ has the form $Bx - Ay + k = 0$.

The given line is $x – 7y + 5 = 0$, so $A=1$ and $B=–7$.

The equation of the perpendicular line is $–7x - 1y + k = 0$, which simplifies to $7x + y - k = 0$. Let $k'=-k$.

$7x + y + k' = 0$.

This line has an x-intercept of 3, so it passes through (3, 0). Substitute these values:

$7(3) + 0 + k' = 0 \implies 21 + k' = 0 \implies k' = -21$.

Substitute $k'=-21$ back into the equation.

$7x + y - 21 = 0$

---

The equation of the required line is $7x + y - 21 = 0$.

Given:

The equations of the two lines are:

Line 1: $\sqrt{3}x + y = 1$

Line 2: $x + \sqrt{3}y = 1$

---

To Find:

The angles between the given lines.

---

Solution:

To find the angle between two lines, we first find their slopes. Let $\theta$ be the acute angle between the lines. The formula relating the angle to the slopes ($m_1$ and $m_2$) is:

$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

Step 1: Find the slopes of the lines.

We convert each equation to the slope-intercept form, $y = mx + c$.

For Line 1: $\sqrt{3}x + y = 1$

$y = -\sqrt{3}x + 1$

The slope of the first line is $m_1 = -\sqrt{3}$.

For Line 2: $x + \sqrt{3}y = 1$

$\sqrt{3}y = -x + 1$

$y = -\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}$

The slope of the second line is $m_2 = -\frac{1}{\sqrt{3}}$.

Step 2: Apply the angle formula.

$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{-\sqrt{3} - (-\frac{1}{\sqrt{3}})}{1 + (-\sqrt{3})(-\frac{1}{\sqrt{3}})} \right|$

$\tan \theta = \left| \frac{-\sqrt{3} + \frac{1}{\sqrt{3}}}{1 + 1} \right| = \left| \frac{\frac{-3 + 1}{\sqrt{3}}}{2} \right|$

$\tan \theta = \left| \frac{-2/\sqrt{3}}{2} \right| = \left| -\frac{1}{\sqrt{3}} \right| = \frac{1}{\sqrt{3}}$

Step 3: Find the angles.

The acute angle $\theta$ for which $\tan \theta = \frac{1}{\sqrt{3}}$ is $30^\circ$.

The other angle between the lines is the obtuse angle, which is given by $180^\circ - \theta$.

$180^\circ - 30^\circ = 150^\circ$.

---

Therefore, the angles between the lines are $30^\circ$ and $150^\circ$.

Given:

A line, $L_1$, passes through the points P(h, 3) and Q(4, 1).

A second line, $L_2$, has the equation $7x - 9y - 19 = 0$.

The lines $L_1$ and $L_2$ are perpendicular to each other.

---

To Find:

The value of h.

---

Solution:

The condition for two lines to be perpendicular is that the product of their slopes is -1.

Step 1: Find the slope of Line 1 ($m_1$).

The slope of the line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is $\frac{y_2 - y_1}{x_2 - x_1}$.

Using the points (h, 3) and (4, 1):

$m_1 = \frac{1 - 3}{4 - h} = \frac{-2}{4 - h}$

Step 2: Find the slope of Line 2 ($m_2$).

We rearrange the equation $7x - 9y - 19 = 0$ into the slope-intercept form ($y=mx+c$).

$9y = 7x - 19$

$y = \frac{7}{9}x - \frac{19}{9}$

The slope of this line is $m_2 = \frac{7}{9}$.

Step 3: Apply the condition for perpendicular lines.

$m_1 \times m_2 = -1$

$\left( \frac{-2}{4 - h} \right) \times \left( \frac{7}{9} \right) = -1$

$\frac{-14}{9(4 - h)} = -1$

Multiply both sides by $-9(4-h)$:

$14 = 9(4 - h)$

$14 = 36 - 9h$

$9h = 36 - 14$

$9h = 22$

$h = \frac{22}{9}$

---

The value of h is $\frac{22}{9}$.

Given:

A point P($x_1, y_1$).

A line $L_1$ with equation $Ax + By + C = 0$.

---

To Prove:

The equation of a line $L_2$ that passes through P and is parallel to $L_1$ is $A(x – x_1) + B(y – y_1) = 0$.

---

Proof:

Method 1: Using Slopes

First, find the slope of the given line $L_1: Ax + By + C = 0$. Assuming $B \neq 0$:

$By = -Ax - C \implies y = -\frac{A}{B}x - \frac{C}{B}$

The slope of $L_1$ is $m_1 = -\frac{A}{B}$.

Since the required line $L_2$ is parallel to $L_1$, its slope $m_2$ is the same: $m_2 = -\frac{A}{B}$.

The line $L_2$ passes through the point $(x_1, y_1)$. Using the point-slope form, $y - y_1 = m_2(x - x_1)$:

$y - y_1 = -\frac{A}{B}(x - x_1)$

Multiply both sides by B:

$B(y - y_1) = -A(x - x_1)$

Rearrange the terms to get the desired form:

$A(x - x_1) + B(y - y_1) = 0$

(If $B=0$, the line is vertical ($x=-C/A$), the parallel line is also vertical ($x=x_1$), and the formula $A(x-x_1)+0(y-y_1)=0$ still holds.)

---

Method 2: Using the Form of Parallel Lines

The equation of any line parallel to $Ax + By + C = 0$ is of the form $Ax + By + k = 0$ for some constant $k$.

This line must pass through the point $(x_1, y_1)$, so these coordinates must satisfy the equation:

$Ax_1 + By_1 + k = 0$

Solving for $k$, we get:

$k = -Ax_1 - By_1$

Now, substitute this expression for $k$ back into the equation of the parallel line:

$Ax + By + (-Ax_1 - By_1) = 0$

$Ax + By - Ax_1 - By_1 = 0$

Group the terms with A and B:

$(Ax - Ax_1) + (By - By_1) = 0$

$A(x - x_1) + B(y - y_1) = 0$

Hence, Proved.

Given:

Two lines intersect at the point P(2, 3).

The angle $\theta$ between the two lines is $60^\circ$.

The slope of one line ($L_1$) is $m_1 = 2$.

---

To Find:

The equation of the other line ($L_2$).

---

Solution:

Let the slope of the other line ($L_2$) be $m$.

The formula for the tangent of the angle $\theta$ between two lines with slopes $m_1$ and $m$ is:

$\tan \theta = \left| \frac{m_1 - m}{1 + m_1 m} \right|$

We are given $\theta = 60^\circ$ and $m_1 = 2$. We know that $\tan 60^\circ = \sqrt{3}$.

Substituting the values:

$\sqrt{3} = \left| \frac{2 - m}{1 + 2m} \right|$

This absolute value equation leads to two possible cases, which will give us two possible slopes for the second line.

---

Case 1: $\frac{2 - m}{1 + 2m} = \sqrt{3}$

$2 - m = \sqrt{3}(1 + 2m)$

$2 - m = \sqrt{3} + 2\sqrt{3}m$

$2 - \sqrt{3} = m + 2\sqrt{3}m$

$2 - \sqrt{3} = m(1 + 2\sqrt{3})$

$m = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}}$

To simplify, we rationalize the denominator:

$m = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} \times \frac{1 - 2\sqrt{3}}{1 - 2\sqrt{3}} = \frac{2 - 4\sqrt{3} - \sqrt{3} + 6}{1^2 - (2\sqrt{3})^2} = \frac{8 - 5\sqrt{3}}{1 - 12} = \frac{8 - 5\sqrt{3}}{-11} = \frac{5\sqrt{3} - 8}{11}$

The equation of the line passing through (2, 3) with this slope is found using the point-slope form $y - y_1 = m(x - x_1)$:

$y - 3 = \left(\frac{5\sqrt{3} - 8}{11}\right)(x - 2)$

$11(y - 3) = (5\sqrt{3} - 8)(x - 2)$

$11y - 33 = (5\sqrt{3} - 8)x - 10\sqrt{3} + 16$

$(5\sqrt{3} - 8)x - 11y + 49 - 10\sqrt{3} = 0$

---

Case 2: $\frac{2 - m}{1 + 2m} = -\sqrt{3}$

$2 - m = -\sqrt{3}(1 + 2m)$

$2 - m = -\sqrt{3} - 2\sqrt{3}m$

$2 + \sqrt{3} = m - 2\sqrt{3}m$

$2 + \sqrt{3} = m(1 - 2\sqrt{3})$

$m = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}$

To simplify, we rationalize the denominator:

$m = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \times \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}} = \frac{2 + 4\sqrt{3} + \sqrt{3} + 6}{1^2 - (2\sqrt{3})^2} = \frac{8 + 5\sqrt{3}}{1 - 12} = -\frac{8 + 5\sqrt{3}}{11}$

The equation of the line passing through (2, 3) with this slope is:

$y - 3 = \left(-\frac{8 + 5\sqrt{3}}{11}\right)(x - 2)$

$11(y - 3) = -(8 + 5\sqrt{3})(x - 2)$

$11y - 33 = -(8 + 5\sqrt{3})x + 16 + 10\sqrt{3}$

$(8 + 5\sqrt{3})x + 11y - 49 - 10\sqrt{3} = 0$

---

Therefore, there are two possible equations for the other line:

$(5\sqrt{3} - 8)x - 11y + (49 - 10\sqrt{3}) = 0$

and

$(8 + 5\sqrt{3})x + 11y - (49 + 10\sqrt{3}) = 0$

Given:

A line segment with endpoints A(3, 4) and B(–1, 2).

---

To Find:

The equation of the right bisector of the segment AB.

---

Solution:

A right bisector is a line that is perpendicular to a segment and passes through its midpoint.

Step 1: Find the midpoint of the line segment AB.

Let the midpoint be M. Using the midpoint formula:

$M = \left( \frac{3 + (–1)}{2}, \frac{4 + 2}{2} \right) = \left( \frac{2}{2}, \frac{6}{2} \right) = (1, 3)$

The right bisector passes through this point M(1, 3).

Step 2: Find the slope of the line segment AB ($m_{AB}$).

$m_{AB} = \frac{2 - 4}{–1 - 3} = \frac{-2}{-4} = \frac{1}{2}$

Step 3: Find the slope of the right bisector ($m_{\perp}$).

The right bisector is perpendicular to AB, so its slope is the negative reciprocal of $m_{AB}$.

$m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{1/2} = -2$

Step 4: Find the equation of the right bisector.

We need the equation of a line that passes through M(1, 3) and has a slope of -2. Using the point-slope form, $y - y_1 = m(x - x_1)$:

$y - 3 = -2(x - 1)$

$y - 3 = -2x + 2$

$2x + y - 5 = 0$

---

The equation of the right bisector is $2x + y - 5 = 0$.

Given:

The point P(–1, 3).

The line $L: 3x – 4y – 16 = 0$.

---

To Find:

The coordinates of the foot of the perpendicular from P to L.

---

Solution:

Let the foot of the perpendicular be the point M(h, k). M is the intersection of the line L and the line PM, which is perpendicular to L.

Step 1: Find the slope of the line L.

$3x – 4y – 16 = 0 \implies 4y = 3x - 16 \implies y = \frac{3}{4}x - 4$

The slope of line L is $m_L = \frac{3}{4}$.

Step 2: Find the equation of the line PM.

Line PM is perpendicular to L, so its slope ($m_{PM}$) is the negative reciprocal of $m_L$.

$m_{PM} = -\frac{1}{3/4} = -\frac{4}{3}$.

Line PM passes through P(–1, 3). Using the point-slope form:

$y - 3 = -\frac{4}{3}(x - (–1))$

$3(y - 3) = -4(x + 1)$

$3y - 9 = -4x - 4$

$4x + 3y - 5 = 0$

Step 3: Find the intersection point M(h, k) of the two lines.

We need to solve the system of linear equations:

Multiply equation (i) by 3 and equation (ii) by 4 to eliminate k.

$9h - 12k = 48$

$16h + 12k = 20$

Adding the two new equations:

$25h = 68 \implies h = \frac{68}{25}$

Substitute $h = \frac{68}{25}$ into equation (ii):

$4\left(\frac{68}{25}\right) + 3k = 5$

$\frac{272}{25} + 3k = 5 \implies 3k = 5 - \frac{272}{25} = \frac{125 - 272}{25} = -\frac{147}{25}$

$k = -\frac{147}{75} = -\frac{49}{25}$

---

The coordinates of the foot of the perpendicular are $\left( \frac{68}{25}, -\frac{49}{25} \right)$.

Given:

A line $L$ has the equation $y = mx + c$.

The perpendicular from the origin O(0, 0) to line L meets it at the point P(–1, 2).

---

To Find:

The values of m and c.

---

Solution:

Step 1: Find the slope of the perpendicular line segment OP.

The line segment OP connects the origin O(0, 0) and the point P(–1, 2). Its slope is:

$m_{OP} = \frac{2 - 0}{–1 - 0} = -2$.

Step 2: Find the slope 'm' of the line L.

The line L is perpendicular to the segment OP. Therefore, the product of their slopes is -1.

The slope of line L is 'm'.

$m \times m_{OP} = -1$

$m \times (–2) = -1 \implies m = \frac{1}{2}$

Step 3: Find the value of 'c'.

The line $y = mx + c$ passes through the point P(–1, 2). We can substitute the coordinates of P and the value of m we just found into the equation to solve for c.

$y = \frac{1}{2}x + c$

$2 = \frac{1}{2}(–1) + c$

$2 = -\frac{1}{2} + c$

$c = 2 + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}$

---

The values are $m = \frac{1}{2}$ and $c = \frac{5}{2}$.

Given:

Line 1 ($L_1$): $x \cos \theta - y \sin \theta - k \cos 2\theta = 0$

Line 2 ($L_2$): $x \sec \theta + y \text{cosec} \theta - k = 0$

$p$ = perpendicular distance from the origin (0, 0) to $L_1$.

$q$ = perpendicular distance from the origin (0, 0) to $L_2$.

---

To Prove:

$p^2 + 4q^2 = k^2$

---

Proof:

The perpendicular distance from the origin to a line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.

Step 1: Calculate p.

For $L_1$, we have $A = \cos \theta$, $B = -\sin \theta$, $C = -k \cos 2\theta$.

$p = \frac{|-k \cos 2\theta|}{\sqrt{\cos^2 \theta + (-\sin \theta)^2}} = \frac{|-k \cos 2\theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = \frac{|-k \cos 2\theta|}{\sqrt{1}} = |k \cos 2\theta|$

Squaring both sides gives:

Step 2: Calculate q.

First, rewrite $L_2$ in terms of sine and cosine:

$\frac{x}{\cos \theta} + \frac{y}{\sin \theta} - k = 0$

Multiply by $\sin\theta \cos\theta$ to clear the denominators:

$x\sin\theta + y\cos\theta - k\sin\theta\cos\theta = 0$

For this line, $A = \sin \theta$, $B = \cos \theta$, $C = -k\sin\theta\cos\theta$.

$q = \frac{|-k\sin\theta\cos\theta|}{\sqrt{\sin^2 \theta + \cos^2 \theta}} = \frac{|-k\sin\theta\cos\theta|}{\sqrt{1}} = |k\sin\theta\cos\theta|$

Using the double angle identity $\sin 2\theta = 2\sin\theta\cos\theta$, we can write $\sin\theta\cos\theta = \frac{1}{2}\sin 2\theta$.

$q = \left| k \cdot \frac{1}{2}\sin 2\theta \right| = \left| \frac{k}{2}\sin 2\theta \right|$

Squaring both sides gives:

$q^2 = \frac{k^2}{4} \sin^2 2\theta \implies 4q^2 = k^2 \sin^2 2\theta$

Step 3: Combine the results.

Add equation (i) and equation (ii):

$p^2 + 4q^2 = k^2 \cos^2 2\theta + k^2 \sin^2 2\theta$

$p^2 + 4q^2 = k^2 (\cos^2 2\theta + \sin^2 2\theta)$

Using the Pythagorean identity $\cos^2\alpha + \sin^2\alpha = 1$ with $\alpha = 2\theta$:

$p^2 + 4q^2 = k^2 (1)$

$p^2 + 4q^2 = k^2$

Hence, Proved.

Given:

Vertices of a triangle are A(2, 3), B(4, –1), and C(1, 2).

---

To Find:

1. The equation of the altitude from vertex A.

2. The length of the altitude from vertex A.

---

Solution:

The altitude from vertex A is the line segment from A that is perpendicular to the opposite side, BC.

Part 1: Equation of the Altitude from A

Step 1: Find the slope of the side BC ($m_{BC}$).

$m_{BC} = \frac{2 - (–1)}{1 - 4} = \frac{3}{-3} = -1$.

Step 2: Find the slope of the altitude from A ($m_{AD}$).

The altitude is perpendicular to BC, so its slope is the negative reciprocal of $m_{BC}$.

$m_{AD} = -\frac{1}{m_{BC}} = -\frac{1}{-1} = 1$.

Step 3: Find the equation of the altitude.

The altitude passes through A(2, 3) and has a slope of 1. Using the point-slope form:

$y - 3 = 1(x - 2)$

$y - 3 = x - 2$

$x - y + 1 = 0$

The equation of the altitude is $x - y + 1 = 0$.

---

Part 2: Length of the Altitude from A

The length of the altitude is the perpendicular distance from point A(2, 3) to the line containing side BC.

Step 1: Find the equation of the line BC.

Using the point-slope form with point C(1, 2) and slope $m_{BC}=-1$:

$y - 2 = -1(x - 1)$

$y - 2 = -x + 1$

$x + y - 3 = 0$

Step 2: Calculate the perpendicular distance from A(2, 3) to the line $x + y - 3 = 0$.

Using the distance formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$:

$d = \frac{|1(2) + 1(3) - 3|}{\sqrt{1^2 + 1^2}} = \frac{|2 + 3 - 3|}{\sqrt{2}} = \frac{|2|}{\sqrt{2}} = \frac{2}{\sqrt{2}}$

Rationalizing the denominator: $d = \frac{2\sqrt{2}}{2} = \sqrt{2}$.

The length of the altitude is $\sqrt{2}$ units.

Given:

A line has an x-intercept 'a' and a y-intercept 'b'.

$p$ is the length of the perpendicular from the origin (0, 0) to this line.

---

To Prove:

$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$

---

Proof:

Step 1: Write the equation of the line in general form.

The equation of a line with intercepts 'a' and 'b' is given by the intercept form:

$\frac{x}{a} + \frac{y}{b} = 1$

To convert this to the general form $Ax + By + C = 0$, we can clear the denominators by multiplying by $ab$:

$bx + ay = ab \implies bx + ay - ab = 0$

Step 2: Use the formula for the perpendicular distance from the origin.

The distance ($p$) from the origin (0, 0) to the line $Ax + By + C = 0$ is:

$p = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}} = \frac{|C|}{\sqrt{A^2 + B^2}}$

For our line, $A=b$, $B=a$, and $C=-ab$.

$p = \frac{|-ab|}{\sqrt{b^2 + a^2}} = \frac{|ab|}{\sqrt{a^2 + b^2}}$

Step 3: Manipulate the equation to get the desired result.

Square both sides of the distance equation:

$p^2 = \frac{a^2 b^2}{a^2 + b^2}$

Take the reciprocal of both sides:

$\frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2}$

Split the fraction on the right-hand side:

$\frac{1}{p^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2}$

Cancel the common terms:

$\frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{a^2}$

Rearranging gives the required relation:

$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$

Hence, Proved.

Given:

The equations of three lines are:

The three lines are concurrent, which means they all intersect at a single common point.

---

To Find:

The value of k.

---

Solution:

The strategy is to find the point of intersection of the two lines whose equations are fully known (L1 and L3), and then substitute this point into the equation of the third line (L2) to find the value of k.

Step 1: Find the point of intersection of L1 and L3.

We need to solve the following system of linear equations:

$2x + y - 3 = 0$

$3x - y - 2 = 0$

We can use the method of elimination. Adding the two equations will eliminate the y-term:

$(2x + y - 3) + (3x - y - 2) = 0$

$5x - 5 = 0$

$5x = 5 \implies x = 1$

Now, substitute $x=1$ into equation (i) to find the value of y:

$2(1) + y - 3 = 0$

$2 + y - 3 = 0$

$y - 1 = 0 \implies y = 1$

The point of intersection of L1 and L3 is (1, 1).

Step 2: Substitute the point of intersection into the equation of L2.

Since the three lines are concurrent, the point (1, 1) must also lie on the line $L_2: 5x + ky - 3 = 0$. Therefore, its coordinates must satisfy this equation.

Substitute $x = 1$ and $y = 1$ into equation (ii):

$5(1) + k(1) - 3 = 0$

$5 + k - 3 = 0$

$2 + k = 0$

$k = -2$

---

Alternate Solution using Determinants:

Three lines $a_1x+b_1y+c_1=0$, $a_2x+b_2y+c_2=0$, and $a_3x+b_3y+c_3=0$ are concurrent if the determinant of their coefficients is zero:

$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$

For the given lines, the coefficients are:

$\begin{vmatrix} 2 & 1 & -3 \\ 5 & k & -3 \\ 3 & -1 & -2 \end{vmatrix} = 0$

Expanding the determinant along the first row:

$2(k(-2) - (-3)(-1)) - 1(5(-2) - (-3)(3)) + (-3)(5(-1) \ $$ - k(3)) = 0$

$2(-2k - 3) - 1(-10 + 9) - 3(-5 - 3k) = 0$

$-4k - 6 - 1(-1) + 15 + 9k = 0$

$-4k - 6 + 1 + 15 + 9k = 0$

$5k + 10 = 0$

$5k = -10$

$k = -2$

---

The value of k is -2.

Given:

A line $L_1: 4x – y = 0$.

A point P(4, 1).

The measurement is taken along a second line, $L_2$, which passes through P and has an angle of inclination of $135^\circ$.

---

To Find:

The specified distance.

---

Solution:

The required distance is the length of the line segment from the point P to the point where the line $L_2$ intersects the line $L_1$. Let's call this intersection point Q.

Step 1: Find the equation of the line $L_2$.

The slope of $L_2$ is $m = \tan(135^\circ) = \tan(180^\circ - 45^\circ) = -\tan(45^\circ) = -1$.

The line $L_2$ passes through P(4, 1) and has a slope of -1. Using the point-slope form, $y - y_1 = m(x - x_1)$:

$y - 1 = -1(x - 4)$

$y - 1 = -x + 4$

$x + y - 5 = 0$. This is the equation of line $L_2$.

Step 2: Find the point of intersection, Q, of lines $L_1$ and $L_2$.

We need to solve the system of equations:

From equation (i), we have $y = 4x$.

Substitute this into equation (ii):

$x + (4x) = 5 \implies 5x = 5 \implies x = 1$.

Now find y: $y = 4(1) = 4$.

The point of intersection is Q(1, 4).

Step 3: Find the distance between P(4, 1) and Q(1, 4).

Using the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:

$d = \sqrt{(1 - 4)^2 + (4 - 1)^2}$

$d = \sqrt{(-3)^2 + 3^2}$

$d = \sqrt{9 + 9} = \sqrt{18}$

$d = \sqrt{9 \times 2} = 3\sqrt{2}$

---

The required distance is $3\sqrt{2}$ units.

Given:

The point P(1, 2).

The line (mirror) $L: x – 3y + 4 = 0$.

---

To Find:

The image of the point P in the line L.

---

Solution:

Let the image of point P(1, 2) be Q(h, k).

The line segment PQ is perpendicular to the mirror line L, and the midpoint of PQ lies on the mirror line L.

Step 1: Use the condition of perpendicularity.

First, find the slope of the mirror line L: $x - 3y + 4 = 0 \implies 3y = x + 4 \implies y = \frac{1}{3}x + \frac{4}{3}$.

The slope of L is $m_L = \frac{1}{3}$.

The slope of the line segment PQ is $m_{PQ} = \frac{k - 2}{h - 1}$.

Since PQ is perpendicular to L, $m_{PQ} \times m_L = -1$.

$\frac{k - 2}{h - 1} \times \frac{1}{3} = -1 \implies k - 2 = -3(h - 1) \implies k - 2 = -3h + 3$

Step 2: Use the midpoint condition.

The midpoint of PQ is $M \left( \frac{1+h}{2}, \frac{2+k}{2} \right)$.

This midpoint M must lie on the mirror line $x - 3y + 4 = 0$.

$\left( \frac{1+h}{2} \right) - 3\left( \frac{2+k}{2} \right) + 4 = 0$

Multiply by 2 to clear the denominators:

$(1+h) - 3(2+k) + 8 = 0$

$1 + h - 6 - 3k + 8 = 0$

$h - 3k + 3 = 0$

Step 3: Solve the system of linear equations.

From equation (i), $k = 5 - 3h$. Substitute this into equation (ii):

$h - 3(5 - 3h) = -3$

$h - 15 + 9h = -3$

$10h = 12 \implies h = \frac{12}{10} = \frac{6}{5}$

Now find k: $k = 5 - 3h = 5 - 3\left(\frac{6}{5}\right) = 5 - \frac{18}{5} = \frac{25 - 18}{5} = \frac{7}{5}$.

---

The image of the point (1, 2) is $\left(\frac{6}{5}, \frac{7}{5}\right)$.

---

Alternate Solution using Formula:

The image $(h, k)$ of a point $(x_1, y_1)$ in the line $ax+by+c=0$ is given by:

$\frac{h - x_1}{a} = \frac{k - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$

Here, $(x_1, y_1)=(1, 2)$ and the line is $x - 3y + 4 = 0$, so $a=1, b=-3, c=4$.

$\frac{h - 1}{1} = \frac{k - 2}{-3} = \frac{-2(1(1) - 3(2) + 4)}{1^2 + (-3)^2} = \frac{-2(1 - 6 + 4)}{1 + 9} = \frac{-2(-1)}{10} = \frac{1}{5}$

Solving for h: $\frac{h - 1}{1} = \frac{1}{5} \implies h = 1 + \frac{1}{5} = \frac{6}{5}$.

Solving for k: $\frac{k - 2}{-3} = \frac{1}{5} \implies k - 2 = -\frac{3}{5} \implies k = 2 - \frac{3}{5} = \frac{7}{5}$.

The image is $\left(\frac{6}{5}, \frac{7}{5}\right)$.

Given:

The equations of three lines forming a triangle are:

$L_1: y = m_1x + c_1$

$L_2: y = m_2x + c_2$

$L_3: x = 0$ (the y-axis)

---

To Prove:

The area of the triangle is $\frac{(c_1 - c_2)^2}{2|m_1 - m_2|}$.

---

Proof:

To find the area of the triangle, we first need to find the coordinates of its three vertices, which are the points of intersection of the lines.

Vertex A (Intersection of $L_1$ and $L_3$):

Substitute $x=0$ into $L_1$: $y = m_1(0) + c_1 = c_1$. So, A = $(0, c_1)$.

Vertex B (Intersection of $L_2$ and $L_3$):

Substitute $x=0$ into $L_2$: $y = m_2(0) + c_2 = c_2$. So, B = $(0, c_2)$.

Vertex C (Intersection of $L_1$ and $L_2$):

Set the y-values equal: $m_1x + c_1 = m_2x + c_2$.

$m_1x - m_2x = c_2 - c_1 \implies (m_1 - m_2)x = c_2 - c_1$.

Assuming $m_1 \neq m_2$, we have $x = \frac{c_2 - c_1}{m_1 - m_2}$.

Substitute this x-value back into $L_1$ to find y:

$y = m_1\left(\frac{c_2 - c_1}{m_1 - m_2}\right) + c_1 = \frac{m_1c_2 - m_1c_1 + c_1(m_1 - m_2)}{m_1 - m_2} = \frac{m_1c_2 - m_1c_1 + m_1c_1 - m_2c_1}{m_1 - m_2} \ $$ = \frac{m_1c_2 - m_2c_1}{m_1 - m_2}$.

So, C = $\left(\frac{c_2 - c_1}{m_1 - m_2}, \frac{m_1c_2 - m_2c_1}{m_1 - m_2}\right)$.

Now, we can calculate the area of triangle ABC. A convenient method is to use the side AB on the y-axis as the base.

Base of the triangle: The length of the base AB is the distance between A(0, $c_1$) and B(0, $c_2$).

Base = $|c_1 - c_2|$.

Height of the triangle: The height corresponding to the base AB is the perpendicular distance from vertex C to the y-axis (the line $x=0$). This distance is simply the absolute value of the x-coordinate of C.

Height = $\left|\frac{c_2 - c_1}{m_1 - m_2}\right| = \frac{|c_2 - c_1|}{|m_1 - m_2|}$.

Area of the triangle:

Area = $\frac{1}{2} \times \text{Base} \times \text{Height}$

Area = $\frac{1}{2} \times |c_1 - c_2| \times \frac{|c_2 - c_1|}{|m_1 - m_2|}$

Since $|c_1 - c_2| = |c_2 - c_1|$, we can write:

Area = $\frac{1}{2} \frac{|c_1 - c_2|^2}{|m_1 - m_2|} = \frac{(c_1 - c_2)^2}{2|m_1 - m_2|}$ (because $|z|^2 = z^2$).

Hence, Proved.

Given:

Line $L_1: 5x – y + 4 = 0$.

Line $L_2: 3x + 4y – 4 = 0$.

A required line, $L$, intersects $L_1$ and $L_2$. The segment of $L$ between $L_1$ and $L_2$ has its midpoint at P(1, 5).

---

To Find:

The equation of the line L.

---

Solution:

Let the required line L pass through P(1, 5) with an unknown slope $m$. Its equation is given by the point-slope form:

$y - 5 = m(x - 1) \implies y = mx - m + 5$.

Let A be the intersection of L and $L_1$, and B be the intersection of L and $L_2$. P(1, 5) is the midpoint of AB.

Step 1: Find the coordinates of A in terms of m.

A is the intersection of $5x - y + 4 = 0$ and $y = mx - m + 5$.

$5x - (mx - m + 5) + 4 = 0$

$5x - mx + m - 5 + 4 = 0$

$x(5 - m) = 1 - m \implies x_A = \frac{1 - m}{5 - m}$

$y_A = 5x_A + 4 = 5\left(\frac{1-m}{5-m}\right) + 4 = \frac{5 - 5m + 20 - 4m}{5-m} = \frac{25 - 9m}{5-m}$.

So, A = $\left(\frac{1-m}{5-m}, \frac{25-9m}{5-m}\right)$.

Step 2: Find the coordinates of B in terms of m.

B is the intersection of $3x + 4y - 4 = 0$ and $y = mx - m + 5$.

$3x + 4(mx - m + 5) - 4 = 0$

$3x + 4mx - 4m + 20 - 4 = 0$

$x(3 + 4m) = 4m - 16 \implies x_B = \frac{4m - 16}{3 + 4m}$.

$y_B = mx_B - m + 5 = m\left(\frac{4m-16}{3+4m}\right) - m + 5 = \frac{4m^2-16m -m(3+4m) + 5(3+4m)}{3+4m} \ $$ = \frac{4m^2-16m -3m-4m^2 + 15+20m}{3+4m} = \frac{m+15}{3+4m}$.

So, B = $\left(\frac{4m - 16}{3 + 4m}, \frac{m+15}{3+4m}\right)$.

Step 3: Use the midpoint formula.

The midpoint of AB is P(1, 5). So, $\frac{x_A + x_B}{2} = 1$.

$\frac{1 - m}{5 - m} + \frac{4m - 16}{3 + 4m} = 2$

$(1 - m)(3 + 4m) + (4m - 16)(5 - m) = 2(5 - m)(3 + 4m)$

$(3 + 4m - 3m - 4m^2) + (20m - 4m^2 - 80 + 16m) = 2(15 + 20m \ $$ - 3m - 4m^2)$

$(3 + m - 4m^2) + (36m - 4m^2 - 80) = 2(15 + 17m - 4m^2)$

$-8m^2 + 37m - 77 = 30 + 34m - 8m^2$

$37m - 77 = 30 + 34m$

$3m = 107 \implies m = \frac{107}{3}$.

Step 4: Find the equation of the line L.

The line passes through P(1, 5) with slope $m = \frac{107}{3}$.

$y - 5 = \frac{107}{3}(x - 1)$

$3(y - 5) = 107(x - 1)$

$3y - 15 = 107x - 107$

$107x - 3y - 92 = 0$

---

The required equation of the line is $107x - 3y - 92 = 0$.

Given:

Two lines:

$L_1: 3x – 2y - 5 = 0$

$L_2: 3x + 2y - 5 = 0$

A moving point P(x, y) is always equidistant from $L_1$ and $L_2$.

---

To Show:

The path (locus) of the point P(x, y) is a straight line.

---

Proof:

The locus of a point equidistant from two intersecting lines is the pair of angle bisectors of the angles between those lines.

The distance from P(x, y) to $L_1$ is $d_1 = \frac{|3x - 2y - 5|}{\sqrt{3^2 + (-2)^2}} = \frac{|3x - 2y - 5|}{\sqrt{13}}$.

The distance from P(x, y) to $L_2$ is $d_2 = \frac{|3x + 2y - 5|}{\sqrt{3^2 + 2^2}} = \frac{|3x + 2y - 5|}{\sqrt{13}}$.

We are given that $d_1 = d_2$.

$\frac{|3x - 2y - 5|}{\sqrt{13}} = \frac{|3x + 2y - 5|}{\sqrt{13}}$

$|3x - 2y - 5| = |3x + 2y - 5|$

This equality holds if and only if:

$3x - 2y - 5 = (3x + 2y - 5)$ OR $3x - 2y - 5 = -(3x + 2y - 5)$

Case 1: $3x - 2y - 5 = 3x + 2y - 5$

$-2y = 2y \implies 4y = 0 \implies y = 0$

The equation $y=0$ represents the x-axis, which is a straight line.

Case 2: $3x - 2y - 5 = -3x - 2y + 5$

$6x = 10 \implies x = \frac{10}{6} \implies x = \frac{5}{3}$

The equation $x = \frac{5}{3}$ represents a vertical line, which is a straight line.

Since the path of the moving point is described by the equations $y=0$ and $x=5/3$, both of which are equations of straight lines, the path is a straight line (specifically, a pair of straight lines).

Hence, Shown.

Given:

The equation of a line is $(k – 3)x – (4 – k^2)y + k^2 – 7k + 6 = 0$.

This is in the general form $Ax + By + C = 0$, where:

$A = k – 3$

$B = –(4 – k^2) = k^2 - 4$

$C = k^2 – 7k + 6$

---

To Find:

The value of k for each of the given conditions.

---

(a) Parallel to the x-axis

A line is parallel to the x-axis if its slope is 0. For a line $Ax+By+C=0$, the slope is $-A/B$. The slope is 0 when $A=0$ and $B \neq 0$.

Set the coefficient of x to zero:

$A = k – 3 = 0 \implies k = 3$.

We must check that for this value of k, the coefficient of y is not zero.

If $k=3$, then $B = k^2 - 4 = 3^2 - 4 = 9 - 4 = 5$.

Since $B \neq 0$, the condition is satisfied.

For the line to be parallel to the x-axis, $k = 3$.

---

(b) Parallel to the y-axis

A line is parallel to the y-axis if its slope is undefined. This occurs when $B=0$ and $A \neq 0$.

Set the coefficient of y to zero:

$B = k^2 - 4 = 0 \implies (k-2)(k+2) = 0$.

This gives two possible values: $k = 2$ or $k = -2$.

We must check that for these values of k, the coefficient of x is not zero.

If $k=2$, then $A = k - 3 = 2 - 3 = -1$. Since $A \neq 0$, $k=2$ is a valid solution.

If $k=-2$, then $A = k - 3 = -2 - 3 = -5$. Since $A \neq 0$, $k=-2$ is a valid solution.

For the line to be parallel to the y-axis, $k = 2$ or $k = -2$.

---

(c) Passing through the origin

A line passes through the origin (0, 0) if its constant term is zero.

Set the constant term C to zero:

$C = k^2 – 7k + 6 = 0$.

We can factor this quadratic equation:

$(k - 1)(k - 6) = 0$.

This gives two possible values: $k = 1$ or $k = 6$.

For the line to pass through the origin, $k = 1$ or $k = 6$.

Given:

The equation of a line is $\sqrt{3}x + y + 2 = 0$.

The normal form of this line is $x \cos \theta + y \sin \theta = p$.

---

To Find:

The values of $\theta$ and $p$.

---

Solution:

The normal form of a line, $x \cos \theta + y \sin \theta = p$, requires that $p$ (the perpendicular distance from the origin) is non-negative ($p \ge 0$).

Step 1: Rewrite the given equation to match the form $Ax+By=C$ with $C>0$.

The given equation is $\sqrt{3}x + y + 2 = 0$.

$\sqrt{3}x + y = -2$.

Since the constant term on the right is negative, we must multiply the entire equation by -1 to make it positive.

$-\sqrt{3}x - y = 2$.

Step 2: Normalize the equation.

We divide the equation by $\sqrt{A^2 + B^2}$, where $A = -\sqrt{3}$ and $B = -1$.

$\sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.

Divide the equation $-\sqrt{3}x - y = 2$ by 2:

$-\frac{\sqrt{3}}{2}x - \frac{1}{2}y = 1$.

Step 3: Compare with the normal form and find $\theta$ and $p$.

Comparing our result with $x \cos \theta + y \sin \theta = p$, we have:

$\cos \theta = -\frac{\sqrt{3}}{2}$

$\sin \theta = -\frac{1}{2}$

$p = 1$

Since both $\cos \theta$ and $\sin \theta$ are negative, the angle $\theta$ lies in the third quadrant.

The reference angle for which $\cos \alpha = \frac{\sqrt{3}}{2}$ and $\sin \alpha = \frac{1}{2}$ is $\alpha = 30^\circ$ or $\frac{\pi}{6}$ radians.

For the third quadrant, $\theta = 180^\circ + 30^\circ = 210^\circ$.

In radians, $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.

---

The required values are $p = 1$ and $\theta = 210^\circ$ (or $\frac{7\pi}{6}$).

Given:

Let the x-intercept be 'a' and the y-intercept be 'b'.

Sum of intercepts: $a + b = 1$.

Product of intercepts: $ab = –6$.

---

To Find:

The equations of the possible lines.

---

Solution:

We need to find pairs of numbers (a, b) that satisfy the given conditions. We can form a quadratic equation whose roots are 'a' and 'b'. A quadratic equation with roots $\alpha$ and $\beta$ is $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.

Here, the sum of roots is 1 and the product is -6. So the quadratic equation for the intercepts is:

$t^2 - (1)t + (-6) = 0 \implies t^2 - t - 6 = 0$.

Factoring the quadratic equation:

$(t - 3)(t + 2) = 0$.

The two roots are $t=3$ and $t=-2$. These are our values for the intercepts 'a' and 'b'. This gives us two possible cases for the pair (a, b).

---

Case 1: x-intercept $a=3$ and y-intercept $b=-2$.

The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.

$\frac{x}{3} + \frac{y}{-2} = 1$.

Multiplying by 6 to clear the denominators:

$2x - 3y = 6 \implies 2x - 3y - 6 = 0$.

---

Case 2: x-intercept $a=-2$ and y-intercept $b=3$.

The equation of the line is:

$\frac{x}{-2} + \frac{y}{3} = 1$.

Multiplying by 6:

$-3x + 2y = 6 \implies 3x - 2y + 6 = 0$.

---

The two possible equations for the lines are $2x - 3y - 6 = 0$ and $3x - 2y + 6 = 0$.

Given:

A line with the equation $\frac{x}{3} + \frac{y}{4} = 1$.

The distance from the line to the required points is 4 units.

The required points lie on the y-axis.

---

To Find:

The coordinates of these points.

---

Solution:

Step 1: Write the equation of the line in the general form $Ax+By+C=0$.

Multiply $\frac{x}{3} + \frac{y}{4} = 1$ by the LCM of 3 and 4, which is 12:

$4x + 3y = 12 \implies 4x + 3y - 12 = 0$.

Step 2: Define a general point on the y-axis.

Any point on the y-axis has an x-coordinate of 0. Let such a point be P(0, b).

Step 3: Use the distance formula.

The distance (d) from a point $(x_1, y_1)$ to the line $Ax+By+C=0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.

Here, $A=4, B=3, C=-12$, the point is $(0, b)$, and the distance $d=4$.

$4 = \frac{|4(0) + 3(b) - 12|}{\sqrt{4^2 + 3^2}}$

$4 = \frac{|3b - 12|}{\sqrt{16 + 9}} = \frac{|3b - 12|}{5}$

$20 = |3b - 12|$.

Step 4: Solve the absolute value equation for b.

This gives two possibilities:

Case 1: $3b - 12 = 20 \implies 3b = 32 \implies b = \frac{32}{3}$.

Case 2: $3b - 12 = -20 \implies 3b = -8 \implies b = -\frac{8}{3}$.

---

The required points on the y-axis are $\left(0, \frac{32}{3}\right)$ and $\left(0, -\frac{8}{3}\right)$.

Given:

A line passes through the points $A(\cos \theta, \sin \theta)$ and $B(\cos \phi, \sin \phi)$.

---

To Find:

The perpendicular distance from the origin (0, 0) to this line.

---

Solution:

Step 1: Find the equation of the line passing through A and B.

Using the two-point form $y - y_1 = m(x - x_1)$, where $m = \frac{y_2-y_1}{x_2-x_1}$:

$y - \sin \theta = \left(\frac{\sin \phi - \sin \theta}{\cos \phi - \cos \theta}\right)(x - \cos \theta)$

$(y - \sin \theta)(\cos \phi - \cos \theta) = (x - \cos \theta)(\sin \phi - \sin \theta)$

$y(\cos \phi - \cos \theta) - \sin\theta\cos\phi + \sin\theta\cos\theta = x(\sin \phi - \sin \theta) - \cos\theta\sin\phi \ $$ + \cos\theta\sin\theta$

$x(\sin \phi - \sin \theta) - y(\cos \phi - \cos \theta) - \cos\theta\sin\phi + \sin\theta\cos\phi = 0$

Using the identity $\sin A \cos B - \cos A \sin B = \sin(A-B)$, the constant term is $\sin(\theta - \phi)$.

The equation is: $x(\sin \phi - \sin \theta) + y(\cos \theta - \cos \phi) + \sin(\theta - \phi) = 0$.

Step 2: Use the formula for the distance from the origin.

The distance (p) from the origin to the line $Ax + By + C = 0$ is $p = \frac{|C|}{\sqrt{A^2 + B^2}}$.

$C = \sin(\theta - \phi) = -\sin(\phi - \theta) = -2\sin\left(\frac{\phi-\theta}{2}\right)\cos\left(\frac{\phi-\theta}{2}\right)$.

$A^2+B^2 = (\sin\phi-\sin\theta)^2 + (\cos\theta-\cos\phi)^2$

$= \sin^2\phi - 2\sin\phi\sin\theta + \sin^2\theta + \cos^2\theta - 2\cos\theta\cos\phi + \cos^2\phi$

$= (\sin^2\phi+\cos^2\phi) + (\sin^2\theta+\cos^2\theta) - 2(\cos\phi\cos\theta + \sin\phi\sin\theta)$

$= 1 + 1 - 2\cos(\phi-\theta) = 2(1 - \cos(\phi-\theta))$

Using the half-angle identity $1-\cos(2\alpha) = 2\sin^2\alpha$, we get:

$A^2+B^2 = 2\left(2\sin^2\left(\frac{\phi-\theta}{2}\right)\right) = 4\sin^2\left(\frac{\phi-\theta}{2}\right)$.

So, $\sqrt{A^2+B^2} = \sqrt{4\sin^2\left(\frac{\phi-\theta}{2}\right)} = \left|2\sin\left(\frac{\phi-\theta}{2}\right)\right|$.

Now, calculate the distance:

$p = \frac{\left|-2\sin\left(\frac{\phi-\theta}{2}\right)\cos\left(\frac{\phi-\theta}{2}\right)\right|}{\left|2\sin\left(\frac{\phi-\theta}{2}\right)\right|} = \left|\cos\left(\frac{\phi-\theta}{2}\right)\right|$.

---

Alternate (Simpler) Method:

The points A and B lie on the unit circle. The line passing through them is a chord. The normal to this chord from the origin bisects the angle between the position vectors of A and B. The angle of the normal is $\omega = \frac{\theta+\phi}{2}$.

The normal form of the line is $x\cos\omega + y\sin\omega = p$.

Since point A lies on this line, it must satisfy the equation:

$(\cos\theta)\cos\left(\frac{\theta+\phi}{2}\right) + (\sin\theta)\sin\left(\frac{\theta+\phi}{2}\right) = p$.

Using the identity $\cos(X-Y) = \cos X \cos Y + \sin X \sin Y$:

$p = \cos\left(\theta - \frac{\theta+\phi}{2}\right) = \cos\left(\frac{2\theta - \theta - \phi}{2}\right) = \cos\left(\frac{\theta-\phi}{2}\right)$.

Since distance must be positive, the perpendicular distance is $|p| = \left|\cos\left(\frac{\theta-\phi}{2}\right)\right|$.

---

The perpendicular distance is $\left|\cos\left(\frac{\theta-\phi}{2}\right)\right|$.

Given:

Line 1: $x – 7y + 5 = 0$.

Line 2: $3x + y = 0$.

The required line is parallel to the y-axis.

---

To Find:

The equation of the line that passes through the intersection of Line 1 and Line 2 and is parallel to the y-axis.

---

Solution:

Step 1: Find the point of intersection of the two given lines.

We have the system of equations:

From equation (ii), it is easy to express y in terms of x: $y = -3x$.

Substitute this into equation (i):

$x - 7(-3x) = -5$

$x + 21x = -5 \implies 22x = -5 \implies x = -\frac{5}{22}$.

We only need the x-coordinate for this problem, but for completeness, $y = -3(-\frac{5}{22}) = \frac{15}{22}$.

The point of intersection is $\left(-\frac{5}{22}, \frac{15}{22}\right)$.

Step 2: Write the equation of the line parallel to the y-axis.

A line parallel to the y-axis is a vertical line. The equation of a vertical line is always of the form $x = k$, where $k$ is a constant equal to the x-coordinate of every point on the line.

Since the required line must pass through the point of intersection, its x-coordinate must be the same as the x-coordinate of the intersection point.

Therefore, the equation of the line is $x = -\frac{5}{22}$.

Rewriting in the general form:

$22x = -5 \implies 22x + 5 = 0$.

---

The required equation of the line is $22x + 5 = 0$.

Given:

A line $L_1$ with the equation $\frac{x}{4} + \frac{y}{6} = 1$.

A required line, $L_2$, is perpendicular to $L_1$.

$L_2$ passes through the point where $L_1$ intersects the y-axis.

---

To Find:

The equation of the line $L_2$.

---

Solution:

Step 1: Find the point where $L_1$ meets the y-axis.

A line meets the y-axis where its x-coordinate is 0. The equation $\frac{x}{a} + \frac{y}{b} = 1$ is the intercept form, where 'b' is the y-intercept.

From the equation $\frac{x}{4} + \frac{y}{6} = 1$, we can directly see that the y-intercept is 6.

Therefore, the line $L_1$ meets the y-axis at the point P(0, 6). The required line $L_2$ passes through this point.

Step 2: Find the slope of the given line $L_1$.

We convert the equation to slope-intercept form ($y=mx+c$).

$\frac{y}{6} = -\frac{x}{4} + 1 \implies y = 6\left(-\frac{x}{4} + 1\right) = -\frac{6}{4}x + 6 = -\frac{3}{2}x + 6$.

The slope of $L_1$ is $m_1 = -\frac{3}{2}$.

Step 3: Find the slope of the required line $L_2$.

Since $L_2$ is perpendicular to $L_1$, its slope ($m_2$) is the negative reciprocal of $m_1$.

$m_2 = -\frac{1}{m_1} = -\frac{1}{-3/2} = \frac{2}{3}$.

Step 4: Find the equation of the required line $L_2$.

$L_2$ passes through the point P(0, 6) and has a slope of $\frac{2}{3}$. Since we have the y-intercept (c=6) and the slope, we can directly use the slope-intercept form.

$y = m_2 x + c \implies y = \frac{2}{3}x + 6$.

To write this in general form, multiply by 3:

$3y = 2x + 18 \implies 2x - 3y + 18 = 0$.

---

The equation of the required line is $2x - 3y + 18 = 0$.

Given:

The three lines forming a triangle are:

$L_1: y – x = 0 \implies y = x$

$L_2: x + y = 0 \implies y = -x$

$L_3: x – k = 0 \implies x = k$

---

To Find:

The area of the triangle.

---

Solution:

To find the area, we first find the coordinates of the three vertices of the triangle by finding the intersection points of the lines.

Vertex A (Intersection of $L_1$ and $L_2$):

$y=x$ and $y=-x \implies x = -x \implies 2x=0 \implies x=0$. Then $y=0$.

So, Vertex A is (0, 0).

Vertex B (Intersection of $L_1$ and $L_3$):

$y=x$ and $x=k$. So, $y=k$.

So, Vertex B is (k, k).

Vertex C (Intersection of $L_2$ and $L_3$):

$y=-x$ and $x=k$. So, $y=-k$.

So, Vertex C is (k, -k).

The vertices of the triangle are A(0, 0), B(k, k), and C(k, -k).

We can find the area using the formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$:

Area = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.

Area = $\frac{1}{2} |0(k - (-k)) + k(-k - 0) + k(0 - k)|$

Area = $\frac{1}{2} |0 - k^2 - k^2| = \frac{1}{2} |-2k^2|$.

Since area must be positive and $k^2$ is always non-negative, this simplifies to:

Area = $\frac{1}{2} (2k^2) = k^2$.

---

Alternate Method (Base and Height):

Consider the side BC as the base of the triangle. Both points B(k, k) and C(k, -k) lie on the vertical line $x=k$.

Length of Base BC = $|k - (-k)| = |2k|$.

The height of the triangle is the perpendicular distance from vertex A(0, 0) to the line containing the base BC (which is the line $x=k$).

Height = $|k - 0| = |k|$.

Area = $\frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times |2k| \times |k| = |k^2| = k^2$.

---

The area of the triangle is $k^2$ square units.

Given:

Three lines are given:

The three lines are concurrent (intersect at one point).

---

To Find:

The value of p.

---

Solution:

If three lines are concurrent, the point of intersection of any two of the lines must lie on the third line. We can find the point of intersection of $L_1$ and $L_3$ (since they don't involve p) and then substitute that point into the equation for $L_2$.

Step 1: Find the point of intersection of $L_1$ and $L_3$.

We solve the system of equations:

$3x + y = 2$

$2x - y = 3$

Using the elimination method, we add the two equations to eliminate y:

$(3x + y) + (2x - y) = 2 + 3$

$5x = 5 \implies x = 1$.

Substitute $x=1$ into the first equation to find y:

$3(1) + y = 2 \implies 3 + y = 2 \implies y = -1$.

The point of intersection of $L_1$ and $L_3$ is (1, -1).

Step 2: Substitute the point of intersection into the equation for $L_2$.

Since the lines are concurrent, the point (1, -1) must satisfy the equation of $L_2$: $px + 2y – 3 = 0$.

$p(1) + 2(-1) – 3 = 0$

$p - 2 - 3 = 0$

$p - 5 = 0$

$p = 5$

---

Alternate Solution (Using Determinants):

Three lines $A_1x+B_1y+C_1=0$, $A_2x+B_2y+C_2=0$, and $A_3x+B_3y+C_3=0$ are concurrent if the determinant of their coefficients is zero.

$\begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix} = 0$

For the given lines, we have:

$\begin{vmatrix} 3 & 1 & -2 \\ p & 2 & -3 \\ 2 & -1 & -3 \end{vmatrix} = 0$

Expanding along the first row:

$3(2(-3) - (-3)(-1)) - 1(p(-3) - (-3)(2)) + (-2)(p(-1) \ $$ - 2(2)) = 0$

$3(-6 - 3) - 1(-3p + 6) - 2(-p - 4) = 0$

$3(-9) + 3p - 6 + 2p + 8 = 0$

$-27 + 3p - 6 + 2p + 8 = 0$

$5p - 25 = 0 \implies 5p = 25 \implies p = 5$.

---

The value of p is 5.

Given:

Three lines are concurrent:

$L_1: y = m_1x + c_1$

$L_2: y = m_2x + c_2$

$L_3: y = m_3x + c_3$

---

To Prove:

$m_1(c_2 – c_3) + m_2(c_3 – c_1) + m_3(c_1 – c_2) = 0$.

---

Proof:

Since the lines are concurrent, they all pass through a single point. Let's find the point of intersection of $L_1$ and $L_2$ and then show it lies on $L_3$.

Step 1: Find the point of intersection of $L_1$ and $L_2$.

At the point of intersection, the coordinates are the same, so:

$m_1x + c_1 = m_2x + c_2$

$x(m_1 - m_2) = c_2 - c_1$

Assuming the slopes are distinct ($m_1 \neq m_2$), the x-coordinate of the intersection is:

$x = \frac{c_2 - c_1}{m_1 - m_2}$

Substitute this x-value into the equation for $L_1$ to find the y-coordinate:

$y = m_1\left(\frac{c_2 - c_1}{m_1 - m_2}\right) + c_1 = \frac{m_1c_2 - m_1c_1 + c_1(m_1 - m_2)}{m_1 - m_2} = \frac{m_1c_2 - m_2c_1}{m_1 - m_2}$

Step 2: Substitute the intersection point into the equation for $L_3$.

Because the lines are concurrent, this intersection point must satisfy the equation for $L_3: y = m_3x + c_3$.

$\frac{m_1c_2 - m_2c_1}{m_1 - m_2} = m_3\left(\frac{c_2 - c_1}{m_1 - m_2}\right) + c_3$

Multiply the entire equation by $(m_1 - m_2)$ to clear the denominator:

$m_1c_2 - m_2c_1 = m_3(c_2 - c_1) + c_3(m_1 - m_2)$

$m_1c_2 - m_2c_1 = m_3c_2 - m_3c_1 + c_3m_1 - c_3m_2$

Step 3: Rearrange the equation to the desired form.

Move all terms to one side:

$m_1c_2 - m_2c_1 - m_3c_2 + m_3c_1 - c_3m_1 + c_3m_2 = 0$

Group the terms by $m_1$, $m_2$, and $m_3$:

$m_1(c_2 - c_3) - m_2c_1 + c_3m_2 - m_3c_2 + m_3c_1 = 0$

$m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3c_1 - m_3c_2 = 0$

$m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0$

Hence, Proved.

Given:

The required lines pass through the point P(3, 2).

They make an angle of $45^\circ$ with the line $L_1: x – 2y = 3$.

---

To Find:

The equations of these lines.

---

Solution:

Let the slope of a required line be $m$. The formula for the angle $\theta$ between two lines with slopes $m_1$ and $m$ is:

$\tan \theta = \left|\frac{m_1 - m}{1 + m_1m}\right|$

Step 1: Find the slope of the given line, $m_1$.

$x - 2y = 3 \implies 2y = x - 3 \implies y = \frac{1}{2}x - \frac{3}{2}$.

So, the slope of the given line is $m_1 = \frac{1}{2}$.

Step 2: Use the angle formula to find the possible slopes of the required lines.

We are given $\theta = 45^\circ$, so $\tan 45^\circ = 1$.

$1 = \left|\frac{1/2 - m}{1 + (1/2)m}\right| = \left|\frac{(1-2m)/2}{(2+m)/2}\right| = \left|\frac{1-2m}{2+m}\right|$.

This gives two cases:

Case 1: $\frac{1-2m}{2+m} = 1 \implies 1-2m = 2+m \implies -1 = 3m \implies m = -\frac{1}{3}$.

Case 2: $\frac{1-2m}{2+m} = -1 \implies 1-2m = -(2+m) = -2-m \implies 3 = m$.

The two possible slopes for the required lines are $m = 3$ and $m = -\frac{1}{3}$.

Step 3: Find the equations of the lines.

Both lines pass through the point (3, 2). We use the point-slope form $y - y_1 = m(x - x_1)$.

For the line with slope $m = 3$:

$y - 2 = 3(x - 3) \implies y - 2 = 3x - 9 \implies 3x - y - 7 = 0$.

For the line with slope $m = -\frac{1}{3}$:

$y - 2 = -\frac{1}{3}(x - 3) \implies 3(y - 2) = -(x - 3) \implies 3y - 6 = -x + 3 \ $$ \implies x + 3y - 9 = 0$.

---

The equations of the two lines are $3x - y - 7 = 0$ and $x + 3y - 9 = 0$.

Given:

Two lines, $L_1: 4x + 7y – 3 = 0$ and $L_2: 2x – 3y + 1 = 0$.

The required line passes through the intersection of $L_1$ and $L_2$.

The required line has equal intercepts on the coordinate axes.

---

To Find:

The equation of the required line.

---

Solution:

Method 1: Find the intersection point first.

Step 1: Find the point of intersection of $L_1$ and $L_2$.

Solve the system:

Multiply equation (ii) by 2: $4x - 6y = -2$.

Subtract this from equation (i): $(4x + 7y) - (4x - 6y) = 3 - (-2) \implies 13y = 5 \implies y = \frac{5}{13}$.

Substitute $y = 5/13$ into equation (ii): $2x - 3(5/13) = -1 \implies 2x = -1 + 15/13 = 2/13 \implies x = \frac{1}{13}$.

The intersection point is P$(\frac{1}{13}, \frac{5}{13})$.

Step 2: Find the equation of the line through P with equal intercepts.

A line with equal non-zero intercepts 'a' has the equation $\frac{x}{a} + \frac{y}{a} = 1$, which simplifies to $x+y=a$.

Since the line passes through P, its coordinates must satisfy the equation:

$\frac{1}{13} + \frac{5}{13} = a \implies a = \frac{6}{13}$.

The equation is $x + y = \frac{6}{13}$, or $13x + 13y = 6 \implies 13x + 13y - 6 = 0$.

---

Method 2: Using the family of lines.

The equation of any line passing through the intersection of $L_1$ and $L_2$ is given by:

$(4x + 7y – 3) + k(2x – 3y + 1) = 0$ for some constant k.

$(4+2k)x + (7-3k)y + (k-3) = 0$.

The x-intercept is found by setting y=0: $x = \frac{-(k-3)}{4+2k} = \frac{3-k}{4+2k}$.

The y-intercept is found by setting x=0: $y = \frac{-(k-3)}{7-3k} = \frac{3-k}{7-3k}$.

Since the intercepts are equal:

$\frac{3-k}{4+2k} = \frac{3-k}{7-3k}$.

One solution is $3-k=0 \implies k=3$.

If $k \neq 3$, then we must have $4+2k = 7-3k \implies 5k = 3 \implies k = 3/5$.

If $k=3$, the equation is $(4+6)x + (7-9)y + (3-3)=0 \implies 10x - 2y = 0 \implies 5x-y=0$. This line passes through the origin, so its intercepts are both 0 (and are equal).

If $k=3/5$, the equation is $(4 + 6/5)x + (7 - 9/5)y + (3/5 - 3) = 0$.

$\frac{26}{5}x + \frac{26}{5}y - \frac{12}{5} = 0 \implies 26x + 26y - 12 = 0 \implies 13x + 13y - 6 = 0$.

Since the point of intersection is not the origin, the case with non-zero intercepts is generally the intended one.

---

The required equation of the line is $13x + 13y - 6 = 0$. (The line $5x-y=0$ is also a valid solution under a broader interpretation).

Given:

A line $L_1$ with equation $y=mx+c$.

A required line $L_2$ passes through the origin (0, 0).

The angle between $L_1$ and $L_2$ is $\theta$.

---

To Prove:

The equation of line $L_2$ is $\frac{y}{x} = \frac{m \pm \tan \theta}{1 \mp m \tan \theta}$.

---

Proof:

The slope of the given line $L_1$ is $m_1 = m$.

Any line passing through the origin has an equation of the form $y = m_2 x$, where $m_2$ is its slope. For any point on this line (other than the origin), we have $m_2 = \frac{y}{x}$.

The formula for the tangent of the angle between two lines is:

$\tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|$.

Substituting the slopes $m_1=m$ and $m_2$:

$\tan \theta = \left|\frac{m_2 - m}{1 + m m_2}\right|$.

This absolute value equation gives two possibilities:

Case 1: $\frac{m_2 - m}{1 + m m_2} = \tan \theta$

$m_2 - m = \tan\theta(1 + mm_2) = \tan\theta + m m_2 \tan\theta$

$m_2 - m m_2 \tan\theta = m + \tan\theta$

$m_2(1 - m\tan\theta) = m + \tan\theta \implies m_2 = \frac{m + \tan\theta}{1 - m\tan\theta}$.

Case 2: $\frac{m_2 - m}{1 + m m_2} = -\tan \theta$

$m_2 - m = -\tan\theta(1 + mm_2) = -\tan\theta - m m_2 \tan\theta$

$m_2 + m m_2 \tan\theta = m - \tan\theta$

$m_2(1 + m\tan\theta) = m - \tan\theta \implies m_2 = \frac{m - \tan\theta}{1 + m\tan\theta}$.

Combining these two results for the slope $m_2$ using the $\pm$ and $\mp$ symbols, we get:

$m_2 = \frac{m \pm \tan\theta}{1 \mp m\tan\theta}$.

Since the equation of the line is $y = m_2 x$, we can write $m_2 = \frac{y}{x}$. Therefore,

$\frac{y}{x} = \frac{m \pm \tan\theta}{1 \mp m\tan\theta}$.

Hence, Proved.

Given:

A line segment joins the points A(–1, 1) and B(5, 7).

A line with the equation $x + y = 4$ divides this segment.

---

To Find:

The ratio in which the segment is divided.

---

Solution:

Method 1: Using Section Formula

Let the line $x+y=4$ divide the segment AB in the ratio $k:1$ at a point P.

Using the section formula, the coordinates of P are:

$P = \left( \frac{k(5) + 1(-1)}{k+1}, \frac{k(7) + 1(1)}{k+1} \right) = \left( \frac{5k-1}{k+1}, \frac{7k+1}{k+1} \right)$.

Since the point P lies on the line $x+y=4$, its coordinates must satisfy this equation.

$\left(\frac{5k-1}{k+1}\right) + \left(\frac{7k+1}{k+1}\right) = 4$

$\frac{5k-1 + 7k+1}{k+1} = 4 \implies \frac{12k}{k+1} = 4$.

$12k = 4(k+1) = 4k + 4$.

$8k = 4 \implies k = \frac{4}{8} = \frac{1}{2}$.

The ratio is $k:1 = \frac{1}{2}:1$. Multiplying by 2 gives the integer ratio $1:2$.

---

Method 2: Using Distances

The ratio in which the line $Ax+By+C=0$ divides the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

Ratio = $-\left(\frac{Ax_1 + By_1 + C}{Ax_2 + By_2 + C}\right)$.

Here, the line is $x+y-4=0$, so $A=1, B=1, C=-4$.

The points are $(x_1, y_1) = (–1, 1)$ and $(x_2, y_2) = (5, 7)$.

Ratio = $-\left(\frac{1(-1) + 1(1) - 4}{1(5) + 1(7) - 4}\right) = -\left(\frac{-1 + 1 - 4}{5 + 7 - 4}\right) = -\left(\frac{-4}{8}\right) = \frac{4}{8} = \frac{1}{2}$.

The ratio is $\frac{1}{2}$, which is written as $1:2$.

(Since the ratio is positive, the division is internal.)

---

The line divides the segment in the ratio 1:2.

Given:

The starting point is P(1, 2).

The target line is $L_1: 4x + 7y + 5 = 0$.

The distance is measured along the line $L_2: 2x – y = 0$.

---

To Find:

The distance from the point P to the line $L_1$ measured along the line $L_2$.

---

Solution:

The problem asks for the distance between the starting point P(1, 2) and the point where the path line intersects the target line $L_1$. The path is specified as being "along the line $2x – y = 0$".

First, we must check if the point P(1, 2) lies on the path line $L_2: 2x – y = 0$.

Substituting the coordinates of P into the equation for $L_2$:

$2(1) - (2) = 2 - 2 = 0$.

Since the equation holds true, the point P(1, 2) lies on the line $2x – y = 0$. Therefore, the path of measurement is the line $L_2$ itself.

The required distance is the distance between the point P(1, 2) and the point of intersection of the two lines, $L_1$ and $L_2$. Let this point of intersection be Q(x, y).

Step 1: Find the point of intersection, Q.

We need to solve the following system of linear equations:

From equation (ii), we can easily express y in terms of x:

$y = 2x$

Now, substitute this expression for y into equation (i):

$4x + 7(2x) + 5 = 0$

$4x + 14x + 5 = 0$

$18x = -5$

$x = -\frac{5}{18}$

Now find the corresponding y-coordinate using $y = 2x$:

$y = 2\left(-\frac{5}{18}\right) = -\frac{10}{18} = -\frac{5}{9}$

The point of intersection is Q$\left(-\frac{5}{18}, -\frac{5}{9}\right)$.

Step 2: Find the distance between P(1, 2) and Q$\left(-\frac{5}{18}, -\frac{5}{9}\right)$.

Using the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:

$d = \sqrt{\left(-\frac{5}{18} - 1\right)^2 + \left(-\frac{5}{9} - 2\right)^2}$

$d = \sqrt{\left(-\frac{5}{18} - \frac{18}{18}\right)^2 + \left(-\frac{5}{9} - \frac{18}{9}\right)^2}$

$d = \sqrt{\left(-\frac{23}{18}\right)^2 + \left(-\frac{23}{9}\right)^2}$

$d = \sqrt{\frac{23^2}{18^2} + \frac{23^2}{9^2}}$

We can factor out $23^2$ from the square root:

$d = \sqrt{23^2 \left(\frac{1}{18^2} + \frac{1}{9^2}\right)} = 23 \sqrt{\frac{1}{324} + \frac{1}{81}}$

Finding a common denominator:

$d = 23 \sqrt{\frac{1}{324} + \frac{4}{324}} = 23 \sqrt{\frac{5}{324}}$

$d = 23 \frac{\sqrt{5}}{\sqrt{324}} = 23 \frac{\sqrt{5}}{18}$

$d = \frac{23\sqrt{5}}{18}$

---

The required distance is $\frac{23\sqrt{5}}{18}$ units.

Given:

A straight line must be drawn through the point P(–1, 2).

This line intersects the line $L: x + y = 4$ at a point Q.

The distance between the point P and the intersection point Q is 3 units.

---

To Find:

The direction of the straight line PQ.

---

Solution:

Let the point of intersection on the line $x+y=4$ be Q(x, y).

The point Q has two properties:

1. It lies on the line $x + y = 4$.

2. Its distance from the point P(–1, 2) is 3 units.

From property 1, we can write the coordinates of Q in terms of a single variable:

So, the coordinates of Q can be written as $(x, 4-x)$.

From property 2, we use the distance formula between P(–1, 2) and Q(x, 4-x):

Distance PQ = $\sqrt{(x - (-1))^2 + ((4-x) - 2)^2} = 3$

$\sqrt{(x + 1)^2 + (2 - x)^2} = 3$

Squaring both sides of the equation:

$(x + 1)^2 + (2 - x)^2 = 9$

Expand the squared terms:

$(x^2 + 2x + 1) + (4 - 4x + x^2) = 9$

$2x^2 - 2x + 5 = 9$

$2x^2 - 2x - 4 = 0$

Divide the entire equation by 2:

$x^2 - x - 2 = 0$

Factor the quadratic equation:

$(x - 2)(x + 1) = 0$

This gives two possible values for the x-coordinate of Q: $x = 2$ or $x = -1$.

We find the corresponding intersection points and then the direction (slope) for each case.

---

Case 1: The intersection point Q has x = 2.

Using $y = 4 - x$, the y-coordinate is $y = 4 - 2 = 2$.

The intersection point is $Q_1(2, 2)$.

The direction of the line is the slope of the line passing through P(–1, 2) and $Q_1(2, 2)$.

Slope $m_1 = \frac{2 - 2}{2 - (-1)} = \frac{0}{3} = 0$.

A slope of 0 indicates a horizontal line, parallel to the x-axis.

---

Case 2: The intersection point Q has x = -1.

Using $y = 4 - x$, the y-coordinate is $y = 4 - (-1) = 5$.

The intersection point is $Q_2(–1, 5)$.

The direction of the line is the slope of the line passing through P(–1, 2) and $Q_2(–1, 5)$.

Slope $m_2 = \frac{5 - 2}{-1 - (-1)} = \frac{3}{0}$.

The slope is undefined, which indicates a vertical line, parallel to the y-axis.

---

Therefore, there are two possible directions:

1. A horizontal direction with a slope of 0.

2. A vertical direction with an undefined slope.

Given:

The endpoints of the hypotenuse of a right-angled triangle are A(1, 3) and B(–4, 1).

---

To Find:

The equations of the legs of the triangle.

---

Solution:

Let the third vertex, where the right angle is located, be C(h, k). The legs of the triangle are the line segments AC and BC.

Since the angle at C is $90^\circ$, the line segment AC is perpendicular to the line segment BC. The product of their slopes must be -1.

Slope of AC ($m_{AC}$) = $\frac{k - 3}{h - 1}$.

Slope of BC ($m_{BC}$) = $\frac{k - 1}{h - (–4)} = \frac{k - 1}{h + 4}$.

Condition for perpendicularity: $m_{AC} \times m_{BC} = -1$.

$\left(\frac{k - 3}{h - 1}\right) \times \left(\frac{k - 1}{h + 4}\right) = -1$

$(k - 3)(k - 1) = -(h - 1)(h + 4)$

$k^2 - 4k + 3 = -(h^2 + 3h - 4)$

$k^2 - 4k + 3 = -h^2 - 3h + 4$

$h^2 + 3h + k^2 - 4k - 1 = 0$

This equation represents the locus of all possible points C such that $\angle ACB = 90^\circ$. This locus is a circle with the hypotenuse AB as its diameter.

To find the equations of the legs, we need to choose a specific point C on this circle. There are infinitely many such triangles. However, two simple cases are when the legs are parallel to the coordinate axes.

---

Case 1: The legs are parallel to the coordinate axes.

If the legs are parallel to the axes, then the vertex C must have coordinates that align with A and B. There are two possibilities for C:

Possibility 1: C has the same x-coordinate as A and the same y-coordinate as B. So, C = (1, 1).

Possibility 2: C has the same x-coordinate as B and the same y-coordinate as A. So, C = (–4, 3).

Let's find the equations of the legs for C = (1, 1).

Leg 1 (Line AC): Passes through A(1, 3) and C(1, 1). This is a vertical line with the equation $x=1$.

Leg 2 (Line BC): Passes through B(–4, 1) and C(1, 1). This is a horizontal line with the equation $y=1$.

So, one possible pair of equations for the legs is $x-1=0$ and $y-1=0$.

Let's find the equations of the legs for C = (–4, 3).

Leg 1 (Line AC): Passes through A(1, 3) and C(–4, 3). This is a horizontal line with the equation $y=3$.

Leg 2 (Line BC): Passes through B(–4, 1) and C(–4, 3). This is a vertical line with the equation $x=-4$.

So, another possible pair of equations for the legs is $y-3=0$ and $x+4=0$.

---

General Solution

Let one leg, passing through A(1, 3), have a slope 'm'. Its equation is:

$y - 3 = m(x - 1)$

The other leg, passing through B(–4, 1), must be perpendicular. Its slope will be $-1/m$. Its equation is:

$y - 1 = -\frac{1}{m}(x - (–4))$

$y - 1 = -\frac{1}{m}(x + 4)$

This pair of equations represents the legs for any non-zero, finite slope 'm'. The cases where the legs are parallel to the axes correspond to $m=0$ or $m$ being undefined.

---

Since the question asks for "an equation" (implying a single pair), the simplest and most common answer is to assume the legs are parallel to the axes.

A possible set of equations for the legs is $x - 1 = 0$ and $y - 1 = 0$.

Another possible set is $y - 3 = 0$ and $x + 4 = 0$.

Given:

The point to be reflected is P(3, 8).

The line of reflection (mirror) is $L: x + 3y = 7$.

---

To Find:

The image of the point P with respect to the line L.

---

Solution:

Let the image of P(3, 8) be Q(h, k).

The line segment PQ is perpendicular to the mirror line L, and the midpoint of PQ lies on L.

Step 1: Use the condition of perpendicularity.

First, find the slope of the mirror line $L: x + 3y = 7 \implies 3y = -x+7 \implies y = -\frac{1}{3}x + \frac{7}{3}$.

The slope of L is $m_L = -\frac{1}{3}$.

The slope of the segment PQ is $m_{PQ} = \frac{k - 8}{h - 3}$.

For perpendicular lines, $m_L \times m_{PQ} = -1$.

$-\frac{1}{3} \times \frac{k - 8}{h - 3} = -1 \implies k - 8 = 3(h - 3) = 3h - 9 \implies 3h - k = 1$.

Step 2: Use the midpoint condition.

The midpoint of PQ is $M \left( \frac{3+h}{2}, \frac{8+k}{2} \right)$.

This midpoint must lie on the line $x + 3y = 7$.

$\left(\frac{3+h}{2}\right) + 3\left(\frac{8+k}{2}\right) = 7$

Multiply by 2: $(3+h) + 3(8+k) = 14 \implies 3+h+24+3k=14 \ $$ \implies h+3k+27=14$.

Step 3: Solve the system of linear equations.

From equation (i), $k = 3h - 1$. Substitute this into equation (ii):

$h + 3(3h - 1) = -13 \implies h + 9h - 3 = -13 \implies 10h = -10 \ $$ \implies h = -1$.

Now find k: $k = 3(-1) - 1 = -3 - 1 = -4$.

---

Alternate Solution (Using Formula):

The image $(h, k)$ of a point $(x_1, y_1)$ in the line $ax+by+c=0$ is given by:

$\frac{h - x_1}{a} = \frac{k - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$

Here, $(x_1, y_1)=(3, 8)$ and the line is $x + 3y - 7 = 0$, so $a=1, b=3, c=-7$.

$\frac{h - 3}{1} = \frac{k - 8}{3} = \frac{-2(1(3) + 3(8) - 7)}{1^2 + 3^2} = \frac{-2(3 + 24 - 7)}{1 + 9} = \frac{-2(20)}{10} = -4$.

Solving for h: $\frac{h - 3}{1} = -4 \implies h - 3 = -4 \implies h = -1$.

Solving for k: $\frac{k - 8}{3} = -4 \implies k - 8 = -12 \implies k = -4$.

---

The image of the point (3, 8) is (–1, –4).

Given:

Line 1: $L_1: y = 3x + 1$. Slope $m_1 = 3$.

Line 2: $L_2: 2y = x + 3 \implies y = \frac{1}{2}x + \frac{3}{2}$. Slope $m_2 = \frac{1}{2}$.

Line 3: $L_3: y = mx + 4$. Slope is $m$.

Lines $L_1$ and $L_2$ are equally inclined to line $L_3$.

---

To Find:

The value of $m$.

---

Solution:

If two lines are equally inclined to a third line, it means the third line is one of the angle bisectors of the first two lines. The angle between $L_1$ and $L_3$ is the same as the angle between $L_2$ and $L_3$.

Let $\theta$ be the angle. Using the formula $\tan\theta = \left|\frac{M_1 - M_2}{1 + M_1 M_2}\right|$:

Angle between $L_1$ and $L_3$: $\tan\theta = \left|\frac{m - m_1}{1 + m m_1}\right| = \left|\frac{m - 3}{1 + 3m}\right|$.

Angle between $L_2$ and $L_3$: $\tan\theta = \left|\frac{m - m_2}{1 + m m_2}\right| = \left|\frac{m - 1/2}{1 + m/2}\right| = \left|\frac{2m - 1}{2 + m}\right|$.

Setting the tangents of the angles equal:

$\left|\frac{m - 3}{1 + 3m}\right| = \left|\frac{2m - 1}{2 + m}\right|$.

This gives two cases:

Case 1: $\frac{m - 3}{1 + 3m} = \frac{2m - 1}{2 + m}$

$(m-3)(2+m) = (2m-1)(1+3m)$

$m^2 - m - 6 = 6m^2 - m - 1 \implies 5m^2 = -5 \implies m^2 = -1$.

This case has no real solution for $m$.

Case 2: $\frac{m - 3}{1 + 3m} = -\left(\frac{2m - 1}{2 + m}\right) = \frac{1 - 2m}{2 + m}$

$(m-3)(2+m) = (1-2m)(1+3m)$

$m^2 - m - 6 = 1 + m - 6m^2$

$7m^2 - 2m - 7 = 0$.

This is a quadratic equation for $m$. We use the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=7, b=-2, c=-7$.

$m = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(7)(-7)}}{2(7)} = \frac{2 \pm \sqrt{4 + 196}}{14} = \frac{2 \pm \sqrt{200}}{14}$

$m = \frac{2 \pm 10\sqrt{2}}{14} = \frac{1 \pm 5\sqrt{2}}{7}$.

---

The possible values of m are $\frac{1 + 5\sqrt{2}}{7}$ and $\frac{1 - 5\sqrt{2}}{7}$.

Given:

A variable point P(x, y).

Line 1: $L_1: x + y – 5 = 0$.

Line 2: $L_2: 3x – 2y + 7 = 0$.

The sum of the perpendicular distances from P to $L_1$ and $L_2$ is always 10.

---

To Show:

The point P must move on a line (i.e., its locus is a straight line).

---

Proof:

Let $d_1$ be the perpendicular distance from P(x, y) to $L_1$, and $d_2$ be the perpendicular distance from P to $L_2$.

$d_1 = \frac{|x + y - 5|}{\sqrt{1^2 + 1^2}} = \frac{|x + y - 5|}{\sqrt{2}}$

$d_2 = \frac{|3x - 2y + 7|}{\sqrt{3^2 + (-2)^2}} = \frac{|3x - 2y + 7|}{\sqrt{13}}$

We are given that $d_1 + d_2 = 10$.

$\frac{|x + y - 5|}{\sqrt{2}} + \frac{|3x - 2y + 7|}{\sqrt{13}} = 10$.

The expressions inside the absolute value signs, $x+y-5$ and $3x-2y+7$, can be positive or negative depending on where the point P(x, y) is located. This leads to four possible equations for the locus of P:

1. $\frac{x + y - 5}{\sqrt{2}} + \frac{3x - 2y + 7}{\sqrt{13}} = 10$

2. $\frac{-(x + y - 5)}{\sqrt{2}} + \frac{3x - 2y + 7}{\sqrt{13}} = 10$

3. $\frac{x + y - 5}{\sqrt{2}} - \frac{3x - 2y + 7}{\sqrt{13}} = 10$

4. $\frac{-(x + y - 5)}{\sqrt{2}} - \frac{3x - 2y + 7}{\sqrt{13}} = 10$

Let's examine the form of any of these equations. For example, equation 1:

$\sqrt{13}(x + y - 5) + \sqrt{2}(3x - 2y + 7) = 10\sqrt{26}$

$\sqrt{13}x + \sqrt{13}y - 5\sqrt{13} + 3\sqrt{2}x - 2\sqrt{2}y + 7\sqrt{2} = 10\sqrt{26}$

$(\sqrt{13} + 3\sqrt{2})x + (\sqrt{13} - 2\sqrt{2})y + (7\sqrt{2} - 5\sqrt{13} - 10\sqrt{26}) = 0$.

This is a linear equation of the form $Ax + By + C = 0$, where A, B, and C are constants. The coefficients A and B are not both zero. Therefore, this equation represents a straight line.

Similarly, each of the other three cases also results in a linear equation of the form $Ax + By + C = 0$.

The locus of the point P is a set of segments of these four lines, forming a parallelogram. Since each piece of the path lies on a straight line, we have shown that P must move on a line.

Hence, Shown.

Given:

Line 1: $L_1: 9x + 6y – 7 = 0$.

Line 2: $L_2: 3x + 2y + 6 = 0$.

---

To Find:

The equation of the line equidistant from $L_1$ and $L_2$.

---

Solution:

The line equidistant from two parallel lines is another parallel line that lies exactly halfway between them.

Step 1: Verify the lines are parallel and write them with identical x and y coefficients.

The slope of $L_1$ is $-\frac{9}{6} = -\frac{3}{2}$.

The slope of $L_2$ is $-\frac{3}{2}$.

Since the slopes are equal, the lines are parallel.

To use the formula for the midline, we need the coefficients of x and y to be the same. We can multiply the equation of $L_2$ by 3.

$3(3x + 2y + 6) = 0 \implies 9x + 6y + 18 = 0$. Let's call this $L_2'$.

Step 2: Use the formula for the equidistant line.

The line equidistant from $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by:

$Ax + By + \frac{C_1 + C_2}{2} = 0$.

Here, $A=9, B=6, C_1=-7, C_2=18$.

The equation is: $9x + 6y + \frac{-7 + 18}{2} = 0$.

$9x + 6y + \frac{11}{2} = 0$.

To eliminate the fraction, multiply the entire equation by 2:

$18x + 12y + 11 = 0$.

---

The equation of the required line is $18x + 12y + 11 = 0$.

Given:

Incident ray passes through P(1, 2).

The reflection surface is the x-axis.

Reflected ray passes through Q(5, 3).

The point of reflection on the x-axis is A.

---

To Find:

The coordinates of point A.

---

Solution:

Method 1: Using Slopes

Let the coordinates of point A on the x-axis be (a, 0).

The slope of the incident ray (PA) is $m_{PA} = \frac{0-2}{a-1} = \frac{-2}{a-1}$.

The slope of the reflected ray (AQ) is $m_{AQ} = \frac{3-0}{5-a} = \frac{3}{5-a}$.

For reflection from a horizontal line (the x-axis), the slope of the reflected ray is the negative of the slope of the incident ray: $m_{AQ} = -m_{PA}$.

$\frac{3}{5-a} = -\left(\frac{-2}{a-1}\right) = \frac{2}{a-1}$.

Cross-multiplying gives: $3(a-1) = 2(5-a) \implies 3a - 3 = 10 - 2a \implies 5a = 13 \implies a = \frac{13}{5}$.

The coordinates of A are $(\frac{13}{5}, 0)$.

---

Method 2: Using the Image of a Point

The reflected ray appears to originate from the image of the source point P reflected in the mirror (the x-axis).

The image of P(1, 2) in the x-axis is P'(1, -2).

The reflected ray passes through Q(5, 3) and appears to come from P'(1, -2). This means the points P', A, and Q are collinear.

Since A is the point where this line crosses the x-axis, we can find the equation of the line passing through P'(1, -2) and Q(5, 3) and then find its x-intercept.

Slope of line P'Q is $m = \frac{3 - (-2)}{5 - 1} = \frac{5}{4}$.

Using point-slope form with Q(5, 3): $y - 3 = \frac{5}{4}(x - 5)$.

To find the x-intercept (point A), set y=0:

$0 - 3 = \frac{5}{4}(x - 5) \implies -12 = 5(x - 5) \implies -12 = 5x - 25 \ $$ \implies 5x = 13 \implies x = \frac{13}{5}$.

The x-intercept is $(\frac{13}{5}, 0)$.

---

The coordinates of A are $\left(\frac{13}{5}, 0\right)$.

Given:

Two points: $P_1(\sqrt{a^2 - b^2}, 0)$ and $P_2(-\sqrt{a^2 - b^2}, 0)$.

A line $L: \frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1$.

---

To Prove:

The product of the perpendicular distances from $P_1$ and $P_2$ to the line L is $b^2$.

---

Proof:

First, write the equation of the line in the general form $Ax+By+C=0$:

$\frac{\cos\theta}{a}x + \frac{\sin\theta}{b}y - 1 = 0$.

The perpendicular distance from a point $(x_1, y_1)$ to this line is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.

Let $d_1$ be the distance from $P_1$ and $d_2$ be the distance from $P_2$.

$d_1 = \frac{|\frac{\cos\theta}{a}(\sqrt{a^2-b^2}) + \frac{\sin\theta}{b}(0) - 1|}{\sqrt{(\frac{\cos\theta}{a})^2 + (\frac{\sin\theta}{b})^2}} = \frac{|\frac{\sqrt{a^2-b^2}\cos\theta}{a} - 1|}{\sqrt{\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}}}$

$d_2 = \frac{|\frac{\cos\theta}{a}(-\sqrt{a^2-b^2}) + \frac{\sin\theta}{b}(0) - 1|}{\sqrt{\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}}} = \frac{|-\frac{\sqrt{a^2-b^2}\cos\theta}{a} - 1|}{\sqrt{\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}}}$

Now, find the product $d_1 d_2$:

$d_1 d_2 = \frac{\left|\left(\frac{\sqrt{a^2-b^2}\cos\theta}{a} - 1\right)\left(-\frac{\sqrt{a^2-b^2}\cos\theta}{a} - 1\right)\right|}{\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}}$

The numerator is of the form $|(X-Y)(-X-Y)| \ $$ = |-(X-Y)(X+Y)| \ $$ = |-(X^2-Y^2)| \ $$ = |Y^2-X^2|$. Let $Y=1$ and $X=\frac{\sqrt{a^2-b^2}\cos\theta}{a}$.

Numerator: $\left|1^2 - \left(\frac{\sqrt{a^2-b^2}\cos\theta}{a}\right)^2\right| = \left|1 - \frac{(a^2-b^2)\cos^2\theta}{a^2}\right| = \left|\frac{a^2 - a^2\cos^2\theta + b^2\cos^2\theta}{a^2}\right|$

Using $a^2(1-\cos^2\theta) = a^2\sin^2\theta$, the numerator becomes $\left|\frac{a^2\sin^2\theta + b^2\cos^2\theta}{a^2}\right| = \frac{a^2\sin^2\theta + b^2\cos^2\theta}{a^2}$.

Denominator: $\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2} = \frac{b^2\cos^2\theta + a^2\sin^2\theta}{a^2b^2}$.

So, $d_1 d_2 = \frac{\frac{a^2\sin^2\theta + b^2\cos^2\theta}{a^2}}{\frac{b^2\cos^2\theta + a^2\sin^2\theta}{a^2b^2}} = \frac{a^2\sin^2\theta + b^2\cos^2\theta}{a^2} \times \frac{a^2b^2}{b^2\cos^2\theta + a^2\sin^2\theta}$

$d_1 d_2 = \frac{\cancel{a^2}b^2(\cancel{a^2\sin^2\theta + b^2\cos^2\theta})}{\cancel{a^2}(\cancel{b^2\cos^2\theta + a^2\sin^2\theta})} = b^2$.

Hence, Proved.

Given:

The person is at the intersection of lines $L_1: 2x – 3y + 4 = 0$ and $L_2: 3x + 4y – 5 = 0$.

The target path is the line $L_3: 6x – 7y + 8 = 0$.

---

To Find:

The equation of the path the person should follow to reach $L_3$ in the least time.

---

Solution:

The path of least time is the shortest distance, which is the perpendicular path from the starting point to the target line.

Step 1: Find the starting point (the intersection of $L_1$ and $L_2$).

Solve the system:

Multiply (i) by 4 and (ii) by 3: $8x-12y=-16$ and $9x+12y=15$.

Add these two equations: $17x = -1 \implies x = -1/17$.

Substitute x into (ii): $3(-1/17) + 4y = 5 \implies 4y = 5 + 3/17 = 88/17 \implies y = 22/17$.

The starting point is P$(-\frac{1}{17}, \frac{22}{17})$.

Step 2: Find the equation of the path perpendicular to the target line $L_3$.

The slope of the target line $L_3: 6x – 7y + 8 = 0$ is $m_3 = -6/(-7) = 6/7$.

The slope of the required path, $m_p$, must be the negative reciprocal of $m_3$.

$m_p = -1/(6/7) = -7/6$.

The path is the line passing through P$(-\frac{1}{17}, \frac{22}{17})$ with slope $-7/6$. Using the point-slope form:

$y - \frac{22}{17} = -\frac{7}{6}\left(x - \left(-\frac{1}{17}\right)\right) \implies y - \frac{22}{17} = -\frac{7}{6}\left(x + \frac{1}{17}\right)$.

$\frac{17y - 22}{17} = -\frac{7}{6}\left(\frac{17x+1}{17}\right)$.

$6(17y - 22) = -7(17x + 1)$.

$102y - 132 = -119x - 7$.

$119x + 102y - 125 = 0$.

---

The equation of the path the person should follow is $119x + 102y - 125 = 0$.