Question

A and B are the points $(2,0)$ and $(0,2)$ respectively. The coordinates of the point P on the line $2x+3y+1=0$ are

• A. $(7,-5)$ if $|PA-PB|$ is maximum
• B. $(\frac{1}{5},\frac{1}{5})$ if $|PA-PB|$ is maximum
• C. $(7,-5)$ if $|PA-PB|$ is minimum
• D. $(\frac{1}{5},\frac{1}{5})$ if $|PA-PB|$ is minimum

Answer 1

The correct option is A $(7,-5)$ if $|PA-PB|$ is maximum

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

$PB+AB>PA$

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

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

$AB=\sqrt{8}$

$AB=2\sqrt{2}$

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 $P(7,-5)$