Straight Lines: Various Forms of Equations

Equation of Lines Parallel to X-axis and Y-axis

Lines parallel to the coordinate axes (x-axis and y-axis) are the simplest types of lines in the Cartesian plane. Their equations are particularly straightforward because one of the coordinates remains constant for all points lying on such a line.

Line Parallel to the X-axis (Horizontal Line)

A straight line is parallel to the x-axis if and only if it is a horizontal line. This means that the line maintains a constant vertical distance from the x-axis. Consequently, every point $(x, y)$ that lies on such a line has the same y-coordinate, regardless of its x-coordinate.

Let this constant y-coordinate value be denoted by $b$. Then, for any point $P(x, y)$ on this line, its y-coordinate must be equal to $b$. This condition defines the equation of the line.

Equation:

The equation of a line parallel to the x-axis and passing through a point with y-coordinate $b$ is:

$\mathbf{y = b}$

• If $b > 0$, the line is located above the x-axis, at a distance of $b$ units.
• If $b < 0$, the line is located below the x-axis, at a distance of $|b|$ units.
• If $b = 0$, the equation becomes $y = 0$. This is the equation of the x-axis itself, which is parallel to itself.

The slope of a horizontal line $y=b$ is $0$. Using the general form $0 \cdot x + 1 \cdot y - b = 0$, the slope $m = -\frac{A}{B} = -\frac{0}{1} = 0$. This is consistent with the angle of inclination being $0^\circ$, as $\tan 0^\circ = 0$.

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Line Parallel to the Y-axis (Vertical Line)

A straight line is parallel to the y-axis if and only if it is a vertical line. This means that the line maintains a constant horizontal distance from the y-axis. Consequently, every point $(x, y)$ that lies on such a line has the same x-coordinate, regardless of its y-coordinate.

Let this constant x-coordinate value be denoted by $a$. Then, for any point $P(x, y)$ on this line, its x-coordinate must be equal to $a$. This condition defines the equation of the line.

Equation:

The equation of a line parallel to the y-axis and passing through a point with x-coordinate $a$ is:

$\mathbf{x = a}$

• If $a > 0$, the line is located to the right of the y-axis, at a distance of $a$ units.
• If $a < 0$, the line is located to the left of the y-axis, at a distance of $|a|$ units.
• If $a = 0$, the equation becomes $x = 0$. This is the equation of the y-axis itself, which is parallel to itself.

The slope of a vertical line $x=a$ is undefined. Using the general form $1 \cdot x + 0 \cdot y - a = 0$, the slope $m = -\frac{A}{B} = -\frac{1}{0}$, which is undefined. This is consistent with the angle of inclination being $90^\circ$, as $\tan 90^\circ$ is undefined.

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Point-Slope Form of a Line

The point-slope form is a useful way to represent the equation of a straight line when you are given the slope of the line and the coordinates of one point that lies on the line. This form is applicable to any non-vertical line.

Derivation of the Formula

Let $L$ be a non-vertical straight line in the Cartesian plane. Assume that we know the slope of this line, denoted by $m$. Let $P_1(x_1, y_1)$ be a specific fixed point that lies on the line $L$.

Now, let $P(x, y)$ be any other arbitrary point on the line $L$. Since $P(x, y)$ is on the line $L$ and is distinct from $P_1(x_1, y_1)$, we can use the definition of slope (using two points) to express the slope of the line segment $P_1P$.

The slope of the line passing through $P_1(x_1, y_1)$ and $P(x, y)$ is given by:

Equation (i) is valid as long as $x \neq x_1$. This is true for any point $P(x, y)$ on the line $L$ that is different from $P_1(x_1, y_1)$, because the line is non-vertical, ensuring that different points on the line have different x-coordinates unless they are the same point.

Now, multiply both sides of equation (i) by $(x - x_1)$: (Since $x \neq x_1$, $x - x_1 \neq 0$, so we are not multiplying by zero).

Rearranging equation (ii) to put the $y$ term first gives the standard Point-Slope Form:

$\mathbf{y - y_1 = m(x - x_1)}$

This equation is satisfied by the coordinates $(x, y)$ of any point on the line $L$. It is also satisfied by the coordinates of the fixed point $P_1(x_1, y_1)$ itself, as substituting $(x_1, y_1)$ into the equation yields $y_1 - y_1 = m(x_1 - x_1)$, which simplifies to $0 = m \cdot 0$, or $0 = 0$. This is a true statement, so $P_1$ lies on the line represented by this equation.

The point-slope form is particularly useful for writing the equation of a line directly when the slope and a point are known, without needing to first find the y-intercept.

Vertical Lines:

The point-slope form $y - y_1 = m(x - x_1)$ is not applicable to vertical lines because the slope $m$ is undefined. However, the equation of a vertical line is simply $x = x_1$, where $(x_1, y_1)$ is any point on the line.

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Two-Point Form of a Line

The two-point form is a method for finding the equation of a straight line when you are given the coordinates of two distinct points that lie on the line. This form is directly derived from the point-slope form by first calculating the slope using the given two points.

Let $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ be two distinct points that lie on the straight line $L$. We want to find the equation that represents all points $(x, y)$ on this line.

Derivation of the Formula

The first step in finding the equation of the line passing through two points is to determine its slope. If the line is non-vertical (i.e., $x_1 \neq x_2$), the slope $m$ of the line passing through $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is given by the slope formula:

Now that we have the slope $m$ (from equation (i)) and we know a point on the line (we can use either $P_1(x_1, y_1)$ or $P_2(x_2, y_2)$), we can use the Point-Slope Form of the equation of a line, which is $y - y_{\text{point}} = m(x - x_{\text{point}})$.

Let's use the point $P_1(x_1, y_1)$ and the slope $m = \frac{y_2 - y_1}{x_2 - x_1}$. Substituting these into the Point-Slope Form:

Equation (ii) is the Two-Point Form of the equation of a straight line. This equation is satisfied by the coordinates $(x, y)$ of any point on the line passing through $P_1$ and $P_2$.

Alternatively, if we had used the point $P_2(x_2, y_2)$ in the Point-Slope Form, we would get:

Equations (ii) and (iii) are equivalent and represent the same line. They can be algebraically manipulated into other forms like the slope-intercept form or the general form.

Special Case - Vertical Line:

If $x_1 = x_2$, the line passing through $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is a vertical line (assuming $y_1 \neq y_2$, otherwise $P_1$ and $P_2$ are the same point). In this case, the denominator $x_2 - x_1$ in the slope formula is zero, and the slope is undefined. The Two-Point Form formula as written in equation (ii) is not directly applicable. However, the equation of a vertical line where all points have the same x-coordinate $x_1$ (or $x_2$) is simply $\mathbf{x = x_1}$ (or $\mathbf{x = x_2}$).

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Slope-Intercept Form of a Line

The slope-intercept form is one of the most fundamental and widely used forms for representing the equation of a non-vertical straight line in the Cartesian coordinate system. This form directly gives us two key pieces of information about the line: its slope and where it crosses the y-axis.

The two parameters that define this form are:

• The slope of the line, denoted by $m$. This describes the steepness and direction of the line.
• The y-intercept of the line, denoted by $c$. This is the y-coordinate of the point where the line intersects the y-axis. The coordinates of this point are always $(0, c)$.

Derivation of the Formula

We can derive the slope-intercept form using the point-slope form of a line. The point-slope form requires knowing the slope $m$ and the coordinates of any one point $(x_1, y_1)$ on the line, and is given by $y - y_1 = m(x - x_1)$.

In the case of the slope-intercept form, we are given the slope $m$ and the y-intercept $c$. The y-intercept corresponds to a specific point on the line, which is $(0, c)$.

We can use the point $(x_1, y_1) = (0, c)$ in the point-slope form. So, substitute $x_1 = 0$ and $y_1 = c$ into the formula $y - y_1 = m(x - x_1)$:

Simplify equation (i):

Now, rearrange equation (ii) to isolate $y$:

This is the Slope-Intercept Form of the equation of a non-vertical line. Any point $(x, y)$ on the line satisfies this equation. Here, $m$ is the slope and $c$ is the y-intercept.

Vertical Lines:

The slope-intercept form $y = mx + c$ cannot represent a vertical line because the slope $m$ of a vertical line is undefined. The equation of a vertical line is always of the form $x = a$, where $a$ is a constant (the x-intercept). A vertical line has no y-intercept unless it is the y-axis itself ($x=0$), in which case it intersects the y-axis at infinitely many points.

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Intercept Form of a Line

The intercept form is a way to write the equation of a straight line based on where it crosses the coordinate axes. This form is particularly useful when the intercepts are known or when you need to quickly find the intercepts from an equation.

This form utilizes two specific points on the line:

• The x-intercept: This is the x-coordinate of the point where the line crosses the x-axis. At this point, the y-coordinate is always 0. If the x-intercept is $a$, the point is $(a, 0)$.
• The y-intercept: This is the y-coordinate of the point where the line crosses the y-axis. At this point, the x-coordinate is always 0. If the y-intercept is $b$, the point is $(0, b)$.

The intercept form is applicable only to lines that intersect both the x-axis and the y-axis at points other than the origin. This means both the x-intercept $a$ and the y-intercept $b$ must be non-zero ($a \neq 0$ and $b \neq 0$). Lines passing through the origin ($a=0$ and $b=0$), horizontal lines ($y=k$, $k \neq 0$, x-intercept undefined), and vertical lines ($x=a$, $a \neq 0$, y-intercept undefined) cannot be represented in this form.

Derivation of the Formula

Assume a line intersects the x-axis at the point $(a, 0)$ and the y-axis at the point $(0, b)$, where $a \neq 0$ and $b \neq 0$. We can derive the equation of this line using the Two-Point Form, which requires two points on the line. Our two points are $P_1(x_1, y_1) = (a, 0)$ and $P_2(x_2, y_2) = (0, b)$.

The Two-Point Form of the equation of a line is $y - y_1 = \left( \frac{y_2 - y_1}{x_2 - x_1} \right) (x - x_1)$.

Substitute the coordinates of $P_1(a, 0)$ and $P_2(0, b)$ into the formula:

Simplify equation (i):

Multiply both sides of equation (iii) by $a$ to eliminate the denominator (since $a \neq 0$):

Distribute $-b$ on the right side of equation (iv):

Rearrange the terms in equation (v) to bring the $x$ and $y$ terms to the left side and the constant term to the right side:

Finally, divide the entire equation (vi) by $ab$. Since we assumed $a \neq 0$ and $b \neq 0$, $ab \neq 0$, so division by $ab$ is allowed:

Simplify the fractions in equation (vii):

This is the Intercept Form of the equation of a line. Here, $a$ is the x-intercept and $b$ is the y-intercept. The equation is valid only for lines that cut both axes away from the origin.

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Normal Form of a Line

The Normal Form, also sometimes referred to as the Perpendicular Form, of the equation of a straight line expresses the line's position in terms of the length of the perpendicular from the origin to the line and the angle that this perpendicular makes with the positive x-axis.

This form is defined by two parameters:

• $p$: The length of the perpendicular segment dropped from the origin $(0, 0)$ to the line. By convention, $p$ is always taken as a non-negative value ($p \ge 0$).
• $\omega$: The angle measured counterclockwise from the positive direction of the x-axis to the perpendicular segment from the origin to the line. The range of $\omega$ is typically $0^\circ \le \omega < 360^\circ$.

If the line passes through the origin, the perpendicular distance $p$ is $0$. In this case, the angle $\omega$ is usually not uniquely defined, or the form might not be considered (as lines through the origin are better represented by $y=mx$).

Derivation of the Formula

Let $L$ be the straight line whose equation we want to find in normal form. Let $ON$ be the perpendicular segment from the origin $O(0, 0)$ to the line $L$. Let $N$ be the foot of this perpendicular on the line $L$.

Given: The length of the perpendicular segment $ON = p$, and the angle that $ON$ makes with the positive x-axis is $\angle XON = \omega$.

From the definition of trigonometric ratios in terms of coordinates, the coordinates of the point $N$ (the foot of the perpendicular on the line) can be determined. N is located at a distance $p$ from the origin along a ray making an angle $\omega$ with the x-axis. Therefore, the coordinates of N are:

The line $L$ passes through the point $N(p \cos \omega, p \sin \omega)$ and is perpendicular to the segment $ON$. We can use the point-slope form to find the equation of line $L$. We need the slope of $L$ and a point on $L$. We already have the point $N$.

First, find the slope of the segment $ON$. If $\cos \omega \neq 0$ (i.e., $\omega \neq 90^\circ, 270^\circ$), the slope of $ON$, denoted $m_{ON}$, is:

Since the line $L$ is perpendicular to $ON$, the product of their slopes is $-1$, provided neither is vertical. If $m_{ON} \neq 0$ (i.e., $\omega \neq 0^\circ, 180^\circ$), the slope of the line $L$, denoted $m_L$, is the negative reciprocal of $m_{ON}$:

Now, use the Point-Slope form $y - y_1 = m(x - x_1)$ for line $L$, using the point $N(p \cos \omega, p \sin \omega)$ and the slope $m_L = -\frac{\cos \omega}{\sin \omega}$ (valid if $\sin \omega \neq 0$):

Multiply both sides of equation (iv) by $\sin \omega$ (assuming $\sin \omega \neq 0$, i.e., $\omega \neq 0^\circ, 180^\circ$):

Expand both sides of equation (v):

$y \sin \omega - p \sin^2 \omega = -x \cos \omega + p \cos^2 \omega$

Rearrange the terms to group $x$ and $y$ terms on one side and the constant term on the other:

$x \cos \omega + y \sin \omega = p \sin^2 \omega + p \cos^2 \omega$

Using the fundamental trigonometric identity $\sin^2 \omega + \cos^2 \omega = 1$, equation (vi) simplifies to:

This is the Normal Form of the equation of a straight line.

Verification for Special Cases (Axes Parallel Lines):

Let's check if this formula holds even when $\sin \omega = 0$ or $\cos \omega = 0$, which were excluded during the derivation steps.

• If $\omega = 0^\circ$: The normal is along the positive x-axis, $N = (p, 0)$. The line $L$ is vertical, $x = p$.

  Normal form: $x \cos 0^\circ + y \sin 0^\circ = p \implies x(1) + y(0) = p \implies x = p$. (Matches)

• If $\omega = 180^\circ$: The normal is along the negative x-axis, $N = (-p, 0)$. The line $L$ is vertical, $x = -p$.

  Normal form: $x \cos 180^\circ + y \sin 180^\circ = p \implies x(-1) + y(0) = p \implies -x = p \implies x = -p$. (Matches)

• If $\omega = 90^\circ$: The normal is along the positive y-axis, $N = (0, p)$. The line $L$ is horizontal, $y = p$.

  Normal form: $x \cos 90^\circ + y \sin 90^\circ = p \implies x(0) + y(1) = p \implies y = p$. (Matches)

• If $\omega = 270^\circ$: The normal is along the negative y-axis, $N = (0, -p)$. The line $L$ is horizontal, $y = -p$.

  Normal form: $x \cos 270^\circ + y \sin 270^\circ = p \implies x(0) + y(-1) = p \implies -y = p \implies y = -p$. (Matches)

Since the formula holds for all cases, $x \cos \omega + y \sin \omega = p$ is the general normal form for any line in the plane (assuming $p \ge 0$).

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Converting General Form to Normal Form

Any linear equation in the general form $Ax + By + C = 0$ can be converted into the normal form $x \cos \omega + y \sin \omega = p$.

The two forms must represent the same line, so their coefficients must be proportional.

1. Start with the general equation: $Ax + By + C = 0$.

2. Rewrite the equation with the constant term on the right side: $Ax + By = -C$.

3. In the normal form $x \cos \omega + y \sin \omega = p$, the right-hand side $p$ must be non-negative ($p \ge 0$). Adjust the equation from step 2 so that its right-hand side is non-negative.

   • If $-C \ge 0$ (i.e., $C \le 0$), the equation is already in the desired form on the right side: $Ax + By = -C$. Let $A' = A, B' = B, D = -C$.
   • If $-C < 0$ (i.e., $C > 0$), multiply the entire equation $Ax + By = -C$ by -1: $-Ax - By = C$. Now the right side $C$ is positive. Let $A' = -A, B' = -B, D = C$.
   So, we have $A'x + B'y = D$, where $D = |-C| \ge 0$. Note that $\sqrt{A'^2 + B'^2} = \sqrt{(\pm A)^2 + (\pm B)^2} = \sqrt{A^2 + B^2}$.

4. Now, compare the equation $A'x + B'y = D$ with the normal form $x \cos \omega + y \sin \omega = p$. Since they represent the same line, their corresponding coefficients must be proportional. There exists a non-zero constant $k$ such that:

   $\cos \omega = k A'$

   ... (i)

   $\sin \omega = k B'$

   ... (ii)

   $p = k D$

   ... (iii)

5. Use the identity $\cos^2 \omega + \sin^2 \omega = 1$ with equations (i) and (ii):

   $(k A')^2 + (k B')^2 = 1$

   $k^2 (A'^2 + B'^2) = 1$

   $k^2 = \frac{1}{A'^2 + B'^2}$

   $k = \pm \frac{1}{\sqrt{A'^2 + B'^2}}$

   From equation (iii), $p = kD$. Since $p \ge 0$ and $D \ge 0$, the constant $k$ must be non-negative ($k \ge 0$). Therefore, we take the positive root for $k$:

   $k = \frac{1}{\sqrt{A'^2 + B'^2}} = \frac{1}{\sqrt{A^2 + B^2}}$

   ... (iv)

   The value $\sqrt{A^2 + B^2}$ is called the normalization factor.

6. Substitute the value of $k$ from equation (iv) back into equations (i), (ii), and (iii) to find the values of $\cos \omega$, $\sin \omega$, and $p$:

   $\cos \omega = \frac{A'}{\sqrt{A^2 + B^2}}$

   ... (v)

   $\sin \omega = \frac{B'}{\sqrt{A^2 + B^2}}$

   ... (vi)

   $p = \frac{D}{\sqrt{A^2 + B^2}} = \frac{|-C|}{\sqrt{A^2 + B^2}} = \frac{|C|}{\sqrt{A^2 + B^2}}$

   ... (vii)

To convert $Ax + By + C = 0$ to normal form:

First, write it as $Ax + By = -C$. Ensure the RHS is non-negative. If $C > 0$, change signs: $-Ax - By = C$.

Then, divide the entire equation by $\sqrt{A^2 + B^2}$ (or $\sqrt{(-A)^2 + (-B)^2}$ if signs were changed):

$\mathbf{\frac{A}{\pm\sqrt{A^2+B^2}} x + \frac{B}{\pm\sqrt{A^2+B^2}} y = \frac{-C}{\pm\sqrt{A^2+B^2}}}$

Choose the sign of $\sqrt{A^2+B^2}$ in the denominator such that the right-hand side $\frac{-C}{\pm\sqrt{A^2+B^2}}$ is positive. The resulting coefficients of $x$ and $y$ will be $\cos \omega$ and $\sin \omega$ respectively, and the RHS will be $p$.

The distance $p = \frac{|C|}{\sqrt{A^2+B^2}}$ is the perpendicular distance from the origin to the line $Ax+By+C=0$.

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