Question

The coordinates of a point $P$ on the line $3x+2y+10=0$ such that $|PA-PB|$ is maximum where $A$ is $(4,2)$ and $B$ is $(2,4)$, are

• A. $(22,28)$
• B. $(22,-28)$
• C. $(-22,28)$
• D. $(-22,-28)$

Answer 1

The correct option is C $(-22,28)$

$|PA-PB|$ is maximum when $P,A,B$ lies on a line

so equation of line containing $A,B$ is

$y-4=m(x-2)m=\frac{2-4}{4-2}=-1\Rightarrow y-4=-1(x-2)\Rightarrow y+x-6=0$

solving eq(1) $3x+2y+10=0$ and wq(2) $y+x-6=0$ we will get $x=-22,y=28$

so point $P$ will be $(-22,28)$

option C is the correct answer