Question

The coordinates of the point $P$ on the line $2x+3y+1=0$ such that $|PA-PB|$ is maximum, where $A(2,0)$ and $B(0,2)$ is

• A. $(4,-3)$
• B. $(7,-5)$
• C. $(10,-7)$
• D. $(-8,5)$

Answer 1

The correct option is B $(7,-5)$

Given Point $P(x_1,y_1)$ lies on

$2x+3y+1=0$-----(1)

Given points $A(2,0),B(0,2)$

maximum value of $|PA-PB|$ is equal to the $AB$

consider PAB is triangle then

from Triangular inequalities

$PB+AB>PA$

$|PA-PB|<PA$-------(2)

Now $PA=\sqrt{(0-2{)}^2+(2-0{)}^2}$

$PA=\sqrt{8}$

$PA=2\sqrt{2}$

from eq (2)

$2\sqrt{2}>|PA-PB|$ is similar to

$2\sqrt{2}>$eq of line $AB$

on solving above eq we get

$y_1=-1(x_1-2)$

$y_1=-x_1+2$

$x_1+y_1=2$----(3)

Point P lies in line (1)

$2x_1+3y_1=-1$----(4)

from eq (3) and (4)

$x_1=7,y_1=-5$

Point $P(7,-5)$