Question

Consider the points $A(3,4)$ and $B(7,13)$. If '$P$' be a point on the line $y=x$ such that $PA+PB$ is minimum, then '$P$' is

• A. $(\frac{-31}{7},\frac{-31}{7})$
• B. $(\frac{31}{7},\frac{31}{7})$
• C. $(\frac{13}{7},\frac{13}{7})$
• D. $(\frac{23}{7},\frac{23}{7})$

Answer 1

The correct option is B $(\frac{31}{7},\frac{31}{7})$

Given ,

$\begin{array}{c}PA+PB=P{A}^{\prime }+PB\\ y-3=\frac{10}{3}\left(x-y\right)\\ 3y-9=10x-40\\ 3y-10x+31=0\\ x=\frac{31}{1},y=\frac{31}{7}\\ y=x\\ \left(\frac{31}{7},\frac{31}{7}\right)\end{array}$

Then,

Option $B$ is correct answer.