Question

A man starts from the point $P(-3,4)$ and reaches at point $Q(0,1)$ touches x-axis at point $R$ such that $PR+RQ$ is least then position of point $R$ is

• A. $(\frac{3}{5},0)$
• B. $(-3,0)$
• C. $(-\frac{3}{5},0)$
• D. $(-\frac{2}{5},0)$

Answer 1

The correct option is C $(-\frac{3}{5},0)$

As point $R$ lies on x-axis. Let $R\equiv (\alpha ,0)$
Now $PR+RQ$ be least. Consider similar ${△}^{\prime }sAPR$ and $BQR$
$\therefore \frac{AR}{BR}=\frac{PA}{QB}$
$\Rightarrow \frac{-3-\alpha }{\alpha -0}=\frac{4}{1}$
$\Rightarrow 4\alpha =-(3+\alpha ),5\alpha =-3$
$\Rightarrow \alpha =-\frac{3}{5}\therefore R(\alpha ,0)=(-\frac{3}{5},0)$
Hence choice (c) is correct answer.