Question

The point $A(2,3),B(4,-1),C(-1,2)$ are the vertices of $△ABC$. Find the length of perpendicular from $C$ on $AB$ and hence find area of $△ABC$.

Answer 1

Equation of $AB$ is $\frac{y-3}{x-2}=\frac{-4}{2}\Rightarrow 2x+y-7=0$

Formula for perpendicular distance from point to line

$D=\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}$

Length of perpendicular from $C(-1,2)$ to $AB$ with equation $2x+y-7=0$ is equal to

$\frac{|2\times (-1)+2-7|}{\sqrt{2^2+1^2}}=\frac{7}{\sqrt{5}}$ units.

$ar\left(△ABC\right)=\frac{2}{\times }\text{base}\times \text{corresponding height}=\frac{1}{2}\times AB\times \frac{7}{\sqrt{5}}=(\frac{1}{2}\times \sqrt{20}\times \frac{7}{\sqrt{5}})=7$
$\therefore$ Area of $△ABC=7$ sq. units.