Question

A man starts from the point $P(4,-3)$ and reaches the point $Q(1,0)$ touching $y-$axis at $R$ such that $PR+RQ$ is minimum, then the point $R$ is:

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

Answer 1

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


$PR+RQ$ needs to be minimum $\Rightarrow P-R-Q$
Let's take the image of $Q(1,0)$ about $y-$axis $\Rightarrow RQ=R{Q}^{\prime }$, where $Q^{\prime }$ is the image of $Q$

$\Rightarrow (PR+RQ{)}_{\text{min}}=(PR+R{Q}^{\prime }{)}_{\text{min}}\Rightarrow P-R-Q^{\prime }$

$\text{Slope of}PR=\text{Slope of}P{Q}^{\prime }\frac{y+3}{0-4}=\frac{0+3}{-1-4}y=-\frac{3}{5}\Rightarrow R(0,-\frac{3}{5})$