Question

Consider the point $A\equiv (0,1)$ and $B\equiv (2,0).{}^{\prime }{P}^{\prime }$ be a point on the line $4x+3y+9=0.$ Coordinate of the point $P$ such that $|PA-PB|$ is minimum, is

• A. $(\frac{3}{20},-\frac{14}{5})$
• B. $(\frac{3}{20},\frac{14}{5})$
• C. $(-\frac{3}{20},\frac{14}{5})$
• D. $(-\frac{9}{20},-\frac{12}{5})$

Answer 1

The correct option is D $(-\frac{9}{20},-\frac{12}{5})$

Minimum value of $|PA-PB|$ is zero. It can be attained if $PA=PB.$ That means $P$ must lie on the perpendicular bisector of $AB$.

So, equation of perpendicular bisector of $AB$ is

$y-\frac{1}{2}=2(x-1)\text{i.e.}y=2x-\frac{3}{2}\cdots \left(1\right)$
Given equation $4x+3y+9=0\cdots \left(2\right)$

Solving both the equation, we get
$P\equiv (-\frac{9}{20},-\frac{12}{5})$