Question

Let $A=(3,-4),B=(1,2).P=(2k-1,2k+1)$ 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 D none of these

Refer to the figure attached.
$P$ is a point on the line $x-y+2=0$.
Both points $A$ and $B$ lie on the same side of the line $x-y+2=0$
Two points $P$ and $P^{\prime }$ are shown on the line.
The point $P$ is obtained by joining the straight line through $A$ and $B^{\prime }$(image of $B$)
We can observe that for point $P^{\prime }$, $A{P}^{\prime }+P^{\prime }{B}^{\prime }>A{B}^{\prime }$ (triangle inequality)
$\Rightarrow A{P}^{\prime }+P^{\prime }{B}^{\prime }>AP+P{B}^{\prime }$
$\Rightarrow A{P}^{\prime }+P^{\prime }{B}^{\prime }>AP+PB$ ($\because PB=P{B}^{\prime }$)
Hence, $AP+PB$ will be minimum if we find $P$ by joining $A$ and $B^{\prime }$
$B^{\prime }$ is the image of $B(1,2)$ about the line $x-y+2=0$.
Hence, the coordinates of $B^{\prime }$ are
$\frac{x^{\prime }-1}{1}=\frac{y^{\prime }-2}{-1}=\frac{-2(1-2+2)}{1^2+1^2}$
Solving we get $(x^{\prime },y^{\prime })\equiv (0,3)$
Now, slope of $A{B}^{\prime }$ is $\frac{3-(-4)}{0-3}=-\frac{7}{3}$
For point $P(2k-1,2k+1)$, slope of $P{B}^{\prime }$ $=\frac{2k+1-3}{2k-1}=\frac{2k-2}{2k-1}$
Equating slopes of $A{B}^{\prime }$ and $P{B}^{\prime }$, and solving we get $k=\frac{13}{20}$
Hence, option D is correct.