Class 10 NCERT Maths Chapter 7: Coordinate Geometry
1. Introduction
Coordinate Geometry combines algebra and geometry using the Cartesian plane. Every point is represented by an ordered pair \((x, y)\) where \(x\) is the horizontal coordinate and \(y\) is the vertical coordinate.
2. Distance Formula
The distance between two points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) is: \[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
3. Section Formula
If a point \(R\) divides the line segment joining \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) in the ratio \(m:n\), then the coordinates of \(R\) are: \[ R\left( \frac{mx_2 + nx_1}{m + n},\ \frac{my_2 + ny_1}{m + n} \right) \]
4. Midpoint Formula
If \(R\) is the midpoint of \(PQ\), then: \[ R\left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right) \]
5. Area of a Triangle
The area of a triangle with vertices \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\) is: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
6. Example Problems
Example 1: Find the distance between \(A(3, 4)\) and \(B(7, 1)\).
\[
AB = \sqrt{(7 - 3)^2 + (1 - 4)^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\]
Example 2: Find the coordinates of the point dividing the line segment joining \(P(2, 3)\) and \(Q(4, -1)\) in the ratio \(3:1\).
\[
R\left( \frac{3 \cdot 4 + 1 \cdot 2}{3 + 1},\ \frac{3 \cdot (-1) + 1 \cdot 3}{3 + 1} \right) = \left( \frac{14}{4},\ \frac{0}{4} \right) = (3.5, 0)
\]
Example 3: Find the area of a triangle whose vertices are \(A(1, 1)\), \(B(4, 4)\), and \(C(6, 1)\).
\[
\text{Area} = \frac{1}{2} \left| 1(4 - 1) + 4(1 - 1) + 6(1 - 4) \right| = \frac{1}{2} \left| 3 + 0 - 18 \right| = \frac{15}{2} = 7.5
\]