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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 |PAPB| 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,B

If three points P,A,B form a triangle, then according to triangle inequalities, |PAPB|<AB

If they do not form a triangle, i.e they are collinear, |PAPB|=AB

If 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=2

Solving the equations 4x+3y=9 and x+2y=2

4x+3y4(x+2y)=98

4x+3y4x8y=17

5y=17 or y=175

Put y=175 in x+2y=2

x=22y=22×175=2345=10345=245

point of intersection of the lines is (245,175)
therefore P is (245,175).

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