Question

Let $A=(3,-4),B(1,2)$ and $P=(2k-1,2k+1)$ is a variable point such that $PA+PB$ is the minimum. Then $k$ is :

• A. $\frac{7}{9}$
• B. $0$
• C. $\frac{7}{8}$
• D. None of these

Answer 1

The correct option is C $\frac{7}{8}$

Since we need the minimum value of $PA+PB$, point $P$ must lie on the line joining $A$ and $B$.

Equating the slopes, we get $\frac{2k+1-2}{2k-1-1}=\frac{-4-2}{3-1}$

$\Rightarrow \frac{2k-1}{2k-2}=\frac{-6}{2}$

$\Rightarrow 2k-1=-3(2k-2)$

i.e. $2k-1=-6k+6$

$\Rightarrow 8k=7$

$\Rightarrow k=\frac{7}{8}$