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Question

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

A
(4,3)
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B
(7,5)
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C
(10,7)
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D
(8,5)
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Solution

The correct option is B (7,5)
Given Point P(x1,y1) lies on
2x+3y+1=0-----(1)
Given points A(2,0),B(0,2)
maximum value of |PAPB| is equal to the AB
consider PAB is triangle then
from Triangular inequalities
PB+AB>PA
|PAPB|<PA-------(2)
Now PA=(02)2+(20)2
PA=8
PA=22
from eq (2)
22>|PAPB| is similar to
22>eq of line AB
on solving above eq we get
y1=1(x12)
y1=x1+2
x1+y1=2----(3)
Point P lies in line (1)
2x1+3y1=1----(4)
from eq (3) and (4)
x1=7,y1=5
Point P(7,5)

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