Question

In the adjoining figure, $PQR$ is a right triangle, right angled at $Q,X$ and $Y$ are the points on $PQ$ and $QR$ such that $PX:XQ=1:2$ and $QY:YR=2:1$. Prove that $9(P{Y}^2+X{R}^2)=13P{R}^2$.

Answer 1

We have,

$X$ divides $PQ$ in ratio$1:2$.

$\Rightarrow XQ=2PX$

$\Rightarrow XQ=\frac{2}{3}PQ......\left(1\right)$

Also, $Y$ divides QR in ratio $2:1$

$QY=2YR$

$\Rightarrow QY=\frac{2}{3}QR.......\left(2\right)$

In $\Delta PQY,$

$P{Y}^2=P{Q}^2+Q{Y}^2$

$\Rightarrow P{Y}^2=P{Q}^2+{\left(\frac{2}{3}QR\right)}^2$

$\Rightarrow P{Y}^2=P{Q}^2+\frac{4}{9}Q{R}^2$

$\Rightarrow 9P{Y}^2=P{Q}^2+4Q{R}^2.......\left(3\right)$

In $\Delta XQR,$

So,

$X{R}^2=X{Q}^2+Q{R}^2$

$\Rightarrow X{R}^2={\left(\frac{2}{3}PQ\right)}^2+Q{R}^2$

$\Rightarrow X{R}^2=\frac{4P{Q}^2}{9}+Q{R}^2$

$\Rightarrow X{R}^2=4P{Q}^2+9Q{R}^2......\left(4\right)$

On adding ( 3) and ( 4) to

$9P{Y}^2+9X{R}^2=9P{Q}^2+4Q{R}^2+4P{Q}^2+9Q{R}^2$

$\Rightarrow 9(P{Y}^2+X{R}^2)=13(P{Q}^2+Q{R}^2)$

$\Rightarrow 9(P{Y}^2+X{R}^2)=13P{R}^2$

Hence, proved.