Question

If a point $Q$ lies on $x$-axis such that $PQ+QR$ is minimum, where $P\equiv (2,1),R\equiv (4,-2)$, then $PQ:QR$ is equal to

• A. $1:2$
• B. $2:1$
• C. $1:3$
• D. $2:5$

Answer 1

The correct option is A $1:2$

Let $Q(\alpha ,0)$
Given points $P\equiv (2,1)\in Q_1,R\equiv (4,-2)\in Q_4$ lies on opposite side of $x$ axis.



Now,
$PQ+QR$ will be minimum when $P,Q,R$ are collinear.
$\Rightarrow$ slope of $PQ$ = slope of $PR$
$\frac{1-0}{2-\alpha }=\frac{1+2}{2-4}$
$\Rightarrow \alpha =\frac{8}{3}$
Thus $Q(\alpha ,0)\equiv (\frac{8}{3},0)$
Now,by distance formula
$PQ=\sqrt{{(2-\frac{8}{3})}^2+(1-0{)}^2}=\sqrt{\frac{13}{9}}$
$QR=\sqrt{{(4-\frac{8}{3})}^2+(0+2{)}^2}=\sqrt{\frac{52}{9}}$
$PQ:QR=\sqrt{\frac{13}{52}}=\sqrt{\frac{1}{4}}=1:2$