Chapter 7 Coordinate Geometry (Concepts)

The Cartesian Plane and Plotting Points

Coordinate geometry, also known as analytical geometry, is a branch of geometry where algebraic principles are applied to geometric concepts. It provides a systematic way to describe the positions of points, lines, curves, and other geometric shapes using numbers.

The foundation of coordinate geometry is the coordinate system. The most common system is the Cartesian coordinate system (or rectangular coordinate system), named after the French mathematician René Descartes. In this system, every point in a plane can be uniquely identified by an ordered pair of numbers, called its coordinates.

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The Cartesian Coordinate Plane

The Cartesian coordinate plane is formed by two mutually perpendicular straight lines in a plane. These lines are called the coordinate axes.

• The horizontal line is called the x-axis. It is often represented by the letter X'OX, where O is the origin, X represents the positive direction, and X' represents the negative direction.
• The vertical line is called the y-axis. It is often represented by the letter YOY', where O is the origin, Y represents the positive direction, and Y' represents the negative direction.

The point where the x-axis and the y-axis intersect is called the origin. The coordinates of the origin are $(0, 0)$.

The coordinate axes divide the plane into four regions called quadrants. The quadrants are numbered in a counter-clockwise direction, starting from the upper-right region:

• The first quadrant ($Q_1$) is the region where both x and y coordinates are positive ($x > 0, y > 0$).
• The second quadrant ($Q_2$) is the region where the x-coordinate is negative and the y-coordinate is positive ($x < 0, y > 0$).
• The third quadrant ($Q_3$) is the region where both x and y coordinates are negative ($x < 0, y < 0$).
• The fourth quadrant ($Q_4$) is the region where the x-coordinate is positive and the y-coordinate is negative ($x > 0, y < 0$).

Points on the axes have specific coordinate forms:

• Any point lying on the x-axis (except the origin) has a y-coordinate of 0. Its coordinates are of the form $(x, 0)$.
• Any point lying on the y-axis (except the origin) has an x-coordinate of 0. Its coordinates are of the form $(0, y)$.
• The origin is the intersection of both axes, with coordinates $(0, 0)$.

Coordinates of a Point:

The location of any point P in the Cartesian plane is uniquely described by an ordered pair of real numbers $(x, y)$.

• The first number, $x$, is called the x-coordinate or abscissa. It represents the perpendicular distance of the point from the y-axis. The sign of the x-coordinate indicates the direction (positive to the right of the y-axis, negative to the left).
• The second number, $y$, is called the y-coordinate or ordinate. It represents the perpendicular distance of the point from the x-axis. The sign of the y-coordinate indicates the direction (positive above the x-axis, negative below).

The coordinates are always written in the order $(x, y)$. For example, the point (3, 5) is located 3 units to the right of the y-axis and 5 units above the x-axis.

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Plotting a Point on the Cartesian Plane

To plot a point with given coordinates $(x, y)$ on the Cartesian plane:

1. Start at the origin: Begin at the point $(0, 0)$ where the x-axis and y-axis intersect.
2. Move horizontally: Move along the x-axis. If the x-coordinate ($x$) is positive, move $x$ units to the right. If $x$ is negative, move $|x|$ units to the left. If $x$ is zero, stay on the y-axis.
3. Move vertically: From the position reached in Step 2, move parallel to the y-axis. If the y-coordinate ($y$) is positive, move $y$ units upwards. If $y$ is negative, move $|y|$ units downwards. If $y$ is zero, stay on the x-axis.
4. Mark the point: The final position reached is the location of the point with coordinates $(x, y)$. Mark this point.

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Distance Formula

One of the fundamental applications of the Cartesian coordinate system is to calculate the distance between any two points in the plane given their coordinates. The distance formula provides a direct way to compute this length.

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Statement of the Distance Formula

Given two points P and Q in the Cartesian plane with coordinates $P(x_1, y_1)$ and $Q(x_2, y_2)$, the distance between them, denoted as PQ or $d$, is given by the formula:

The order of subtraction in the differences doesn't matter because the differences are squared ($(a-b)^2 = (b-a)^2$). So, the formula can also be written as $d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$.

A special case is finding the distance of a point $P(x, y)$ from the origin $O(0, 0)$. Using the distance formula with $(x_1, y_1) = (0, 0)$ and $(x_2, y_2) = (x, y)$, the distance is:

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Derivation of the Distance Formula

The distance formula is derived using the Pythagorean theorem. Let the two points be $P(x_1, y_1)$ and $Q(x_2, y_2)$. We assume, without loss of generality, that $x_2 \ge x_1$ and $y_2 \ge y_1$ for ease of explanation with a diagram in the first quadrant, but the derivation holds for any points.

Draw a line segment connecting points P and Q. To form a right-angled triangle, draw a line through P parallel to the x-axis and a line through Q parallel to the y-axis. These two lines will intersect at a point, let's call it R.

The coordinates of point R will be $(x_2, y_1)$. Point R has the same x-coordinate as Q and the same y-coordinate as P.

The triangle $\triangle \text{PQR}$ is a right-angled triangle with the right angle at R.

• The length of the horizontal side PR is the distance between the x-coordinates of P and R. Since P is $(x_1, y_1)$ and R is $(x_2, y_1)$, the length $PR = |x_2 - x_1|$.
• The length of the vertical side QR is the distance between the y-coordinates of Q and R. Since Q is $(x_2, y_2)$ and R is $(x_2, y_1)$, the length $QR = |y_2 - y_1|$.
• The distance we want to find, PQ, is the hypotenuse of the right triangle $\triangle \text{PQR}$.

By the Pythagoras Theorem (Chapter 6), in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Substitute the lengths of PR and QR:

Since squaring a real number makes it non-negative, the absolute value signs can be removed:

Take the square root of both sides to find the distance PQ. Since distance is a non-negative value, we take the positive square root:

This is the distance formula, which allows us to calculate the distance between any two points $(x_1, y_1)$ and $(x_2, y_2)$ in the Cartesian plane.

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Section Formula

In coordinate geometry, the section formula is a formula used to determine the coordinates of a point that lies on a line segment and divides it into two parts with a given ratio. This point can divide the segment internally (between the two endpoints) or externally (outside the segment on the line containing it). In Class 10, we primarily focus on internal division.

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Statement of the Section Formula (Internal Division)

Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be two given points in the Cartesian plane. Let $P(x, y)$ be a point that lies on the line segment AB and divides it internally in the ratio $m_1 : m_2$. This means that the ratio of the length of segment AP to the length of segment PB is $m_1/m_2$, i.e., $\frac{\text{AP}}{\text{PB}} = \frac{m_1}{m_2}$.

The coordinates of the point $P(x, y)$ are given by the section formula:

In these formulas, $m_1$ is the ratio corresponding to the segment from the first point (A) to P, and $m_2$ is the ratio corresponding to the segment from the second point (B) to P. The denominator is the sum of the ratios, $m_1 + m_2$.

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Derivation of the Section Formula (Internal Division)

Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be the two given points, and let $P(x, y)$ be the point that divides the line segment $AB$ internally in the ratio $m_1 : m_2$, such that $\frac{\text{AP}}{\text{PB}} = \frac{m_1}{m_2}$.

Draw lines from points A, P, and B perpendicular to the x-axis, meeting the x-axis at points $C(x_1, 0)$, $Q(x, 0)$, and $R(x_2, 0)$ respectively. Similarly, draw lines from A, P, and B perpendicular to the y-axis, meeting the y-axis at points $S(0, y_1)$, $T(0, y)$, and $U(0, y_2)$ respectively.

From the figure, the x-coordinates $x_1, x, x_2$ are on the x-axis. The distance $CQ = x - x_1$ and $QR = x_2 - x$. Similarly, the y-coordinates $y_1, y, y_2$ are on the y-axis. The distance $ST = y - y_1$ and $TU = y_2 - y$.

Now, draw a line through A parallel to the x-axis, intersecting PQ (extended if needed) and BR at points L and M respectively. Draw a line through P parallel to the x-axis, intersecting BR at point N.

Consider $\triangle \text{APL}$ and $\triangle \text{PNB}$.

• $\angle \text{ALP} = 90^\circ$ and $\angle \text{PNB} = 90^\circ$ (Since AL and PN are parallel to the x-axis, and PL and BN are parts of lines perpendicular to the x-axis).
• $\angle \text{PAL} = \angle \text{BPN}$ (These are corresponding angles formed by the transversal APB intersecting the parallel lines AL and PN, which are parallel to the x-axis).

Therefore, $\triangle \text{APL} \sim \triangle \text{PNB}$ by AA Similarity Criterion (Section I3).

Since the triangles are similar, the ratio of their corresponding sides is equal to the ratio of the corresponding parts of the line segment AB:

We are given that $\frac{\text{AP}}{\text{PB}} = \frac{m_1}{m_2}$.

From the coordinates:

• $AL = x - x_1$ (Horizontal distance from $(x_1, y_1)$ to $(x, y_1)$)
• $PN = x_2 - x$ (Horizontal distance from $(x, y_2)$ to $(x_2, y_2)$) - NO, this is incorrect projection.

Let's use the projections onto the axes directly as in the simplified approach mentioned previously.

The ratio of segments on the line AB, $AP/PB = m_1/m_2$, is preserved under projection onto the x-axis. The projection of A is C, P is Q, and B is R. Thus, Q divides CR in the ratio $m_1 : m_2$.

The length $CQ$ is the distance between $(x_1, 0)$ and $(x, 0)$, which is $|x - x_1|$. Since P is between A and B, Q is between C and R (or R and C, depending on the order of $x_1, x_2$). Assuming $x_1 < x < x_2$, $CQ = x - x_1$ and $QR = x_2 - x$.

Cross-multiply equation (1):

Gather terms involving $x$ on one side and other terms on the other side:

Solve for $x$ by dividing by $(m_1 + m_2)$ (which is non-zero as $m_1, m_2$ are ratios of lengths):

Similarly, projecting onto the y-axis, the point T divides the segment SU in the ratio $m_1 : m_2$.

The length $ST = |y - y_1|$ and $TU = |y_2 - y|$. Assuming $y_1 < y < y_2$, $ST = y - y_1$ and $TU = y_2 - y$.

Cross-multiply equation (3):

Gather terms involving $y$ on one side:

Solve for $y$:

Equations (2) and (4) give the coordinates of the point $P(x, y)$ that divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m_1 : m_2$. This completes the derivation.

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Midpoint Formula

A special case of the section formula is when the point P is the midpoint of the line segment AB. The midpoint divides the segment into two equal parts, so the ratio of division is $1 : 1$. This means $m_1 = 1$ and $m_2 = 1$.

We can find the coordinates of the midpoint $M(x, y)$ by substituting $m_1 = 1$ and $m_2 = 1$ into the section formula:

Thus, the coordinates of the midpoint M of the line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ are given by the midpoint formula:

The midpoint's coordinates are simply the average of the x-coordinates and the average of the y-coordinates of the two endpoints.

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Area of a Triangle using Coordinates

In previous geometry studies, you learned how to calculate the area of a triangle using formulas like $\frac{1}{2} \times \text{base} \times \text{height}$ or Heron's formula (using side lengths). In coordinate geometry, if the coordinates of the three vertices of a triangle are known, we can directly calculate its area using a specific formula derived from coordinate principles.

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Statement of the Area Formula using Coordinates

Let the three vertices of a triangle be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$. The area of $\triangle \text{ABC}$ is given by the formula:

The formula involves summing three terms, each being the x-coordinate of a vertex multiplied by the difference of the y-coordinates of the other two vertices (in a cyclic order). The absolute value (denoted by vertical bars $|...|$) is taken because area is always a non-negative quantity. The order of the terms within the sum is cyclic (1, 2, 3 then 2, 3, 1 then 3, 1, 2 for the indices of $x$ and $y$).

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Idea Behind the Derivation of the Area Formula

The formula for the area of a triangle in terms of its vertices' coordinates can be derived using the concept of areas of trapezoids or rectangles formed by projecting the vertices onto one of the coordinate axes (usually the x-axis or y-axis). The area of the triangle is then expressed as the sum or difference of the areas of these simpler figures.

For instance, consider a triangle ABC with vertices $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$. Let's assume, for simplicity in the diagram, that the x-coordinates are ordered $x_1 < x_2 < x_3$ and the y-coordinates are positive. Draw perpendiculars from the vertices A, B, and C to the x-axis, meeting the x-axis at points $D(x_1, 0), E(x_2, 0), F(x_3, 0)$ respectively.

From the figure, the area of $\triangle \text{ABC}$ can be expressed as the sum of the areas of trapezoid ABED and trapezoid BCFE, minus the area of trapezoid ACFD. (Note: The ordering of vertices and the signs of y-coordinates affect whether areas are added or subtracted, but the final formula accounts for this using the absolute value).

• Trapezoid ABED has parallel sides AD (length $|y_1|$) and BE (length $|y_2|$), and height DE (length $|x_2 - x_1|$). Area $= \frac{1}{2}(|y_1| + |y_2|)|x_2 - x_1|$.
• Trapezoid BCFE has parallel sides BE (length $|y_2|$) and CF (length $|y_3|$), and height EF (length $|x_3 - x_2|$). Area $= \frac{1}{2}(|y_2| + |y_3|)|x_3 - x_2|$.
• Trapezoid ACFD has parallel sides AD (length $|y_1|$) and CF (length $|y_3|$), and height DF (length $|x_3 - x_1|$). Area $= \frac{1}{2}(|y_1| + |y_3|)|x_3 - x_1|$.

For the specific case where $y_1, y_2, y_3$ are all positive and $x_1 < x_2 < x_3$, Area($\triangle \text{ABC}$) = Area(Trapezoid ABED) + Area(Trapezoid BCFE) - Area(Trapezoid ACFD)

Substituting the areas and expanding the expression $x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$ and considering all possible orderings of vertices leads to the general formula. The formula can also be derived using vector cross products, which is a more advanced method.

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Condition for Collinearity of Three Points

Three points in a plane are collinear if and only if they lie on the same straight line. Geometrically, this means that if you try to form a triangle with these three points as vertices, the resulting figure will be a degenerate triangle with zero area.

So, three points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ are collinear if and only if the area of the triangle formed by them is zero.

Using the area formula, this condition translates to:

Since $\frac{1}{2}$ is not zero and the absolute value of a number is zero if and only if the number itself is zero, the condition for collinearity simplifies to:

This equation is a useful test for checking if three given points are collinear.

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