Question

Consider two points $A\equiv (-3,0)$ and $B\equiv (0,4).$ A point $P$ on line $2x-3y-12=0$ is such that $|PA-PB|$ is maximum. Then $P$ is

• A. $(-12,-12)$
• B. $(6,0)$
• C. $(0,-4)$
• D. None of these

Answer 1

The correct option is A $(-12,-12)$

$|PA-PB|$ will be maximum if $P,A,B$ are collinear.
Hence, equation of $AB$ is
$\frac{x}{-3}+\frac{y}{4}=1\cdots \left(1\right)$
and given line $2x-3y-12=0\cdots \left(2\right)$
On solving equation $\left(1\right)$ and $\left(2\right),$ we get
$P\equiv (-12,-12)$