Question

A man starts walking from the point $P(-3,4),$ touches the $x-$ axis at $R,$ and then turns to reach at the point $Q(0,2).$ The man is walking at a constant speed. If the man reaches the point $Q$ in the minimum time, then $50(PR{)}^2+(RQ{)}^2$ is equal to

Answer 1

To minimize distance $PR+RQ$
Take mirror image of $P$ in $y=0$
$P^{\prime }=(-3,-4)$
If we join $P^{\prime }Q$ we will get required $R$
Equation of $P^{\prime }Q\Rightarrow y=2x+2$
So $R=(-1,0)$
$P=(-3,4),R(-1,0),Q(0,2)$
$P{R}^2+R{Q}^2=20+5=25$