Question

The co-ordinates of a point 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. $(11,10)$
• C. $(11,17)$
• D. $(0,5)$

Answer 1

Answer: (D)

$(0,5)$

Let $P=(x,2x+5)$.

Hence

$PA=\sqrt{(x-4{)}^2+(2x+3{)}^2}$ ...(i)

$PB=\sqrt{(x-2{)}^2+(2x+1{)}^2}$ ...(ii)

Now

$|PA-PB|$

$=k\left(x\right)$

$=|\sqrt{(x-4{)}^2+(2x+3{)}^2}-\sqrt{(x-2{)}^2+(2x+1{)}^2}|$

Now from the options we can see that there are only two available options for x.

They are $x=11$ and $x=0$.

Now

$k\left(0\right)=5-\sqrt{5}=2.76$ ...(a)

$k\left(11\right)=\sqrt{674}-\sqrt{610}=1.26$ ...(b)

Hence

$|PA-PB|$ is maximum at $x=0$.

Hence $P=(0,5)$.