CBSE · Class 6-12

Class 10 · Mathematics (Basic)

Unit 7 · Class 10Mathematics (Basic)

Coordinate Geometry

Distance formula, section formula, area of a triangle from coordinates.

Lessons

6 concepts

1
Coordinate Plane
The plane is divided into 4 quadrants by the $x$ - and $y$ -axes at the origin. A point $P(x, y)$ has abscissa $x$ (distance from $y$ -axis) and ordinate $y$ (from $x$ -axis). Signs of coordinates identify the quadrant.
2
Distance Formula
Between $A(x_1, y_1)$ and $B(x_2, y_2)$ : $AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ . Derivation uses Pythagoras on the right triangle with horizontal and vertical legs $|x_2 - x_1|$ and $|y_2 - y_1|$ .
3
Section Formula
The point dividing $AB$ internally in ratio $m:n$ has coordinates $P = \left(\dfrac{mx_2 + nx_1}{m+n}, \dfrac{my_2 + ny_1}{m+n}\right)$ . For external division, replace $n$ by $-n$ . Midpoint is $m:n = 1:1$ .
4
Area of a Triangle
Given vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ : area $= \tfrac{1}{2}\big|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\big|$ . If this is $0$ , the three points are collinear.
5
Applications
Use these formulas to verify types of quadrilaterals (parallelogram: diagonals bisect each other; rhombus: all four sides equal; rectangle: diagonals equal and bisecting), or to find centroids and other special points.
6
Distance from Origin & Special Points
The centroid of a triangle with vertices $(x_i, y_i)$ is $G = \left(\dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3}\right)$ . This divides each median in $2:1$ from vertex to midpoint of opposite side.

Skills to practise

Skill · medium

Coordinate Geometry — Practice