Question

Let$P\equiv (\alpha ,\beta )$ be a point on the line $3x+2y+10=0$ such that $|PA-PB|$ is maximum and $A\equiv (4,2)$ , $B\equiv (2,4)$, then the value of $|\alpha |+|\beta |$ is equal to

Answer 1

Given line, $3x+2y+10=0$ $\cdots$(1)
Let $L(x,y)=3x+2y+10$
$L(4,2)=26>0$
$L(2,4)=24>0$
$\therefore A,B$ will lie on same side of $L=0$
$|PA-PB|$ is maximum when $P,A,B$ are collinear
So,$P$ is the Intersection point of line $L=0$ and line $AB$

Equation of $AB$ : $y-2=\frac{2-4}{4-2}(x-4)$
$\Rightarrow x+y-6=0\cdots \left(2\right)$
Solving $\left(1\right),\left(2\right)$ we get, $P\equiv (-22,28)$
$\therefore |\alpha |+|\beta |=50$