A man starts walking from the point P ( - 3, 4 ) , touches the x...

For minimum time, value of $PR + RQ$ must be minimum

so $R$ lies on $P{Q}^{'}$(where ${Q}^{'}$is image of $Q$ w.r.t $x$-axis)

From the diagram we can easily say ${Q}^{'} ( 0, - 2 )$

As we know that \left(y - {y}_{1}\right) = \dfrac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}\left(x - {x}_{1}\right)$</p><p>Equation of$P{Q}^{'}\left(y + 2\right) = \dfrac{4 + 2}{- 3 - 0}\left(x - 0\right)$</p><p>$⇒$Equation of$P{Q}^{'} is $2x + y + 2 = 0$

For the coordinate of point $R$ put $y = 0$

$⇒ R\left(- 1, 0\right)$

$⇒ 50\left(( PR{)}^{2} + ( RQ{)}^{2}\right)$

Distance formula $= \sqrt{{( {x}_{2} - {x}_{1} )}^{2} + {( {y}_{2} - {y}_{1} )}^{2}}$

$⇒ PR = \sqrt{{( - 3 + 1 )}^{2} + {( 4 - 0 )}^{2}} \text{ and } RQ = \sqrt{{( 0 + 1 )}^{2} + {2}^{2}}$

$= 50 ( 20 + 5 ) = 1250$