Coordinate Geometry – Class 9 CBSE Mathematics

1. Introduction

Coordinate geometry, also known as analytic geometry, studies the geometric properties of shapes using a coordinate system. It helps in locating points on a plane using coordinates.

• The plane is divided by two perpendicular lines:

  • x-axis (horizontal)

  • y-axis (vertical)

• Their intersection is called the origin (O), with coordinates $(0,0)$.

2. Coordinates of a Point

A point in the plane is represented by an ordered pair $(x, y)$:

• $x$ = distance from y-axis (abscissa)

• $y$ = distance from x-axis (ordinate)

Example: Point $P(3,4)$ is 3 units to the right of y-axis and 4 units above x-axis.

3. Distance Formula

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

$$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$

4. Midpoint Formula

The midpoint $M$ of a line segment joining $P(x_1, y_1)$ and $Q(x_2, y_2)$ is:

$$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$

5. Section Formula

If a point $P(x, y)$ divides a line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m:n$, then:

$$x = \frac{mx_2 + nx_1}{m+n}, \quad y = \frac{my_2 + ny_1}{m+n}$$

Special Case: Midpoint

When $m = n = 1$, the section formula reduces to the midpoint formula.

6. Slope of a Line

The slope (or gradient) of the line joining points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is:

$$m = \frac{y_2 – y_1}{x_2 – x_1}, \quad x_2 \neq x_1$$

• Positive slope → line rises from left to right

• Negative slope → line falls from left to right

• Zero slope → horizontal line

• Undefined slope → vertical line

7. Equation of a Line

1. Slope-Intercept Form:

$$y = mx + c$$

2. Point-Slope Form:

$$y – y_1 = m(x – x_1)$$

3. Two-Point Form:

$$y – y_1 = \frac{y_2 – y_1}{x_2 – x_1} (x – x_1)$$

8. Distance of a Point from a Line

For a line in the form $Ax + By + C = 0$, the perpendicular distance of a point $P(x_1, y_1)$ from the line is:

$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

9. Collinearity of Points

Three points $P(x_1, y_1), Q(x_2, y_2), R(x_3, y_3)$ are collinear if:

$$\frac{y_2 – y_1}{x_2 – x_1} = \frac{y_3 – y_2}{x_3 – x_2}$$

Or the area of the triangle formed by them is zero:

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

10. Summary Table