Chapter 7

Class 10 Mathematics - Chapter 7: Coordinate Geometry
Chapter 7: Coordinate Geometry
1. Distance Formula
Distance Formula: The distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by: $$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Distance from Origin: The distance of a point $P(x, y)$ from the origin $O(0, 0)$ is: $$OP = \sqrt{x^2 + y^2}$$
P(x₁, y₁) Q(x₂, y₂) X Y
Example 1.1: Calculating Distance Between Two Points

Find the distance between the points $P(2, 3)$ and $Q(4, 1)$.

Solution:
Here $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (4, 1)$.
Using the distance formula: $$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$PQ = \sqrt{(4 - 2)^2 + (1 - 3)^2}$$ $$PQ = \sqrt{(2)^2 + (-2)^2}$$ $$PQ = \sqrt{4 + 4}$$ $$PQ = \sqrt{8} = 2\sqrt{2}\text{ units}$$ The distance is $2\sqrt{2}\text{ units}$.

2. Section Formula
Internal Section Formula: The coordinates of the point $P(x, y)$ which divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m_1 : m_2$ are: $$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$$ $$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$$
Mid-point Formula: The mid-point $M$ of the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ divides it in the ratio $1 : 1$: $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
Example 2.1: Finding Coordinates Using Section Formula

Find the coordinates of the point which divides the line segment joining $(-1, 7)$ and $(4, -3)$ in the ratio $2 : 3$.

Solution:
Let $A(-1, 7)$ and $B(4, -3)$. Here $m_1 = 2$ and $m_2 = 3$.

Applying section formula for $x$: $$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$$ $$x = \frac{2(4) + 3(-1)}{2 + 3}$$ $$x = \frac{8 - 3}{5} = \frac{5}{5} = 1$$
Applying section formula for $y$: $$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$$ $$y = \frac{2(-3) + 3(7)}{2 + 3}$$ $$y = \frac{-6 + 21}{5} = \frac{15}{5} = 3$$
The required point is $(1, 3)$.

Example 2.2: Finding the Division Ratio

In what ratio does the point $(-4, 6)$ divide the line segment joining $A(-6, 10)$ and $B(3, -8)$?

Solution:
Let the ratio be $k : 1$.
Using the section formula for $x$-coordinate: $$x = \frac{k(3) + 1(-6)}{k + 1}$$ $$-4 = \frac{3k - 6}{k + 1}$$ $$-4(k + 1) = 3k - 6$$ $$-4k - 4 = 3k - 6$$ $$-4k - 3k = -6 + 4$$ $$-7k = -2$$ $$k = \frac{2}{7}$$ Thus, the required ratio is $2 : 7$.

3. Area of a Triangle
Formula: The area of $\triangle ABC$ formed by vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ is: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$
Collinearity Condition: Points $A$, $B$, and $C$ are collinear if and only if: $$\text{Area}(\triangle ABC) = 0$$ $$x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$$
Concept Formula Key Condition
Distance $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Always non-negative
Mid-point $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ Ratio is $1:1$
Collinear Points $\text{Area}(\triangle ABC) = 0$ Points lie on a straight line
Example 3.1: Finding Area of Triangle

Find the area of a triangle whose vertices are $(1, -1)$, $(-4, 6)$, and $(-3, -5)$.

Solution:
Let $(x_1, y_1) = (1, -1)$, $(x_2, y_2) = (-4, 6)$, and $(x_3, y_3) = (-3, -5)$.

Substitute into the area formula: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ $$\text{Area} = \frac{1}{2} |1(6 - (-5)) + (-4)((-5) - (-1)) + (-3)((-1) - 6)|$$ $$\text{Area} = \frac{1}{2} |1(11) + (-4)(-4) + (-3)(-7)|$$ $$\text{Area} = \frac{1}{2} |11 + 16 + 21|$$ $$\text{Area} = \frac{1}{2} |48| = 24\text{ sq. units}$$ The area of the triangle is $24\text{ sq. units}$.