Question

Let $P$ be $(5,3)$ and a point $R$ on $y=x$ and $Q$ on the $X$ - axis be such $PQ+QR+RP$ is minimum. Then the coordinates of $Q$ are

• A. $(0,\frac{17}{4})$
• B. $(\frac{17}{4},0)$
• C. $(0,17)$
• D. $(17,0)$

Answer 1

The correct option is B $(\frac{17}{4},0)$

For $PQ+QR+RP$ to be minimum,
$PQ+QR+RP=PQ+QR+R{P}^{\prime }$
$[\because Q,P\text{lies on same side w.r.t}y=x]$
$=PQ+Q{P}^{\prime }$
$=Q{P}^{\prime \prime }+Q{P}^{\prime }$
$=P^{\prime }{P}^{\prime \prime }[\because P,P^{\prime }\text{lies on same side w.r.t}X\text{axis}]$
$\therefore$ Minimum value occurs when $P^{\prime },Q,P^{\prime \prime }$ are collinear.
$\Rightarrow m_{P^{\prime }Q}=m_{P^{\prime \prime }Q}$
$\Rightarrow \frac{5-0}{3-\alpha }=\frac{0+3}{\alpha +5}$
$\Rightarrow \alpha =\frac{17}{4}$
$\therefore Q$ is $(\frac{17}{4},0)$