A man starts walking from the point P ( - 3 ,   4 ) , touches the x -axis at R , and then turns to…

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

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

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

$$As we know that $\left(y-y_1\right)=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

$$Equation of $P{Q}^{\text{'}} \left(y+2\right)=\frac{4+2}{-3-0}\left(x-0\right)$

$\Rightarrow$ Equation of$P{Q}^{\text{'}}$ is $2x+y+2=0$

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

$\Rightarrow R\left(-1, 0\right)$

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

Distance formula $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$\Rightarrow PR=\sqrt{{(-3+1)}^2+{(4-0)}^2}\& RQ=\sqrt{{(0+1)}^2+2^2}$

$=50(20+5)=1250$