Question

$A(0,3)$, $B(-1,-2)$ and $C(4,2)$ are vertices of a $\Delta ABC$. $D$ is a point on the side $BC$ such that $\frac{BD}{DC}=\frac{1}{2}$. $P$ is a point on $AD$ such that $AP=\frac{2\sqrt{5}}{3}$ units. Find coordinates of $P$.

Answer 1

We have the points $B(-1,-2)$ and $C(4,2)$.

Given, $BD:DC=1:2$.

$\therefore$ co-ordinates of $D$ $=(\frac{-2+4}{2+1},\frac{-4+2}{2+1})$$=(\frac{2}{3},-\frac{2}{3})$.

Now, $AD=\sqrt{{(0-\frac{2}{3})}^2+{(3+\frac{2}{3})}^2}=\frac{5\sqrt{5}}{3}$.

Given, $AP=\frac{2\sqrt{5}}{3}$.

$\therefore PD=\frac{3\sqrt{5}}{3}$.

So, $AP:PD=2:3$.

$\therefore$ co-ordinates of $P$ is $\left(\frac{0+\frac{4}{3}}{2+3},\frac{9-\frac{4}{3}}{2+3}\right)$$=\left(\frac{4}{15},\frac{23}{15}\right)$