Question

The point $A$ divides the join of $P(-5,1)$ and $Q(3,5)$ in the ratio $k:1$. Find the two values for $k$ for which the area of triangle ABC, where $B$ is $(1,5)$ and $C$ is $(7,-2)$, is equals to $2$ units.

Answer 1

Formula used: internal division

$B(1,5),C(7,-2),A(\frac{-5+3k}{k+1},\frac{1+5k}{k+1})$

Area of $△ABC=\frac{1}{2}\left|\frac{-5+3k}{k+1}\right(7)+1(-2-\frac{1+5k}{k+1})+7(\frac{1+5k}{k+1}-5)|=2$

$|\frac{-35+21k}{k+1}-2-\left(\frac{1+5k}{k+1}\right)+\frac{7+35k}{k+1}-35|=4$

$|\frac{-35+21k-1-5k+7+35k}{k+1}-37|=4$

$|\frac{-29+51k}{k+1}-37|=4$

$\left|\frac{-29+51k-37k-37}{k+1}\right|=4$

$\left|\frac{14k-66}{k+1}\right|=4$

$\frac{14k-66}{k+1}=\pm 4$

$\frac{14k-66}{k+1}=4\Rightarrow 14k-66=4k+4\Rightarrow k=7$

$\frac{14k-66}{k+1}=-4\Rightarrow 14k-66=-4k-4\Rightarrow k=\frac{31}{9}$