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Question

Consider the points A(0, 1) and B(2, 0), and P be a point on the line 4x+3y+9=0. The coordinates of P such that |PAâˆ’PB| is maximum are

A
(125,15)
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B
(845,135)
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C
(245,175)
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D
(0, 3)
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Solution

The correct option is B (−245,175)Consider P,A,BIf three points P,A,B form a triangle, then according to triangle inequalities, |PA−PB|<AB If they do not form a triangle, i.e they are collinear, |PA−PB|=ABIf they are collinear, then P is the point of intersection of lines AB,4x+3y+9=0 (given P is on this line)equation of line AB=x+2y=2Solving the equations 4x+3y=−9 and x+2y=2⇒4x+3y−4(x+2y)=−9−8⇒4x+3y−4x−8y=−17⇒−5y=−17 or y=175Put y=175 in x+2y=2⇒x=2−2y=2−2×175=2−345=10−345=−245point of intersection of the lines is (−245,175)therefore P is (−245,175).

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