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Question

# The coordinates of the point P on the line 2x+3y+1=0 such that |PAā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 on2x+3y+1=0-----(1)Given points A(2,0),B(0,2)maximum value of |PA−PB| is equal to the ABconsider PAB is triangle thenfrom Triangular inequalities PB+AB>PA|PA−PB|<PA-------(2)Now PA=√(0−2)2+(2−0)2PA=√8PA=2√2from eq (2)2√2>|PA−PB| is similar to 2√2>eq of line ABon solving above eq we get y1=−1(x1−2)y1=−x1+2x1+y1=2----(3)Point P lies in line (1) 2x1+3y1=−1----(4)from eq (3) and (4)x1=7,y1=−5Point P(7,−5)

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