Question

The co-ordinates of a P on the line $2x-y+5=0$ such that |PA-PB| is maximum where A is $(4,-2)$ and B is $(2,-4)$ will be:

• A. $(11,27)$
• B. $\left(-11,-17\right)$
• C. $(-11,17)$
• D. $\left(0,5\right)$

Answer 1

Answer: (B)

$\left(-11,-17\right)$

The maximum value of $|PA-PB|$ is AB.

$AB=\sqrt{(4-2{)}^2+(-2+4{)}^2}=2\sqrt{2}$

Let $p(x,y)$

Then $PA=\sqrt{(x-4{)}^2+(y+2{)}^2}$

$PB=\sqrt{(x-2{)}^2+(y+4{)}^2}$

$(PA-PB)=2\sqrt{2}$

Also $2x-y+5=0$

Put $y=2x+5$

$|PA-PB|=2\sqrt{2}$

$|\sqrt{(x-4{)}^2+(2x+7{)}^2}-\sqrt{(x-2{)}^2+(2x+9{)}^2}|=2\sqrt{2}$

On solving we get $x=-11$

$y=2x+5=-17$

B is correct.