Question

Let $P=(1,1)$ and $Q=(3,2)$, the point $R$ on the x-axis such that $PR+RQ$ is the minimum is :

• A. $(\frac{5}{3},0)$
• B. $(\frac{1}{3},0)$
• C. $(3,0)$
• D. none of these

Answer 1

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

Let us assume point x-axis is $R(x,0)$
So $PR+RQ=\sqrt{(x-1{)}^2+1}+\sqrt{(x-3{)}^2+4}$
we have to calculate minimum distance so we will calculate derivative of this
$f^{\prime }\left(x\right)=\frac{x-1}{\sqrt{(x-1{)}^2+1}}+\frac{x-3}{\sqrt{(x-3{)}^2+4}}=0$
it gives $x=\frac{5}{3}$