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Question

Consider the point A≡(0,1) and B≡(2,0). ′P′ be a point on the line 4x+3y+9=0. Coordinate of the point P such that |PA−PB| is minimum, is

A
(320,145)
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B
(320,145)
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C
(320,145)
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D
(920,125)
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Solution

The correct option is D (−920,−125)Minimum value of |PA−PB| is zero. It can be attained if PA=PB. That means P must lie on the perpendicular bisector of AB. So, equation of perpendicular bisector of AB is y−12=2(x−1)i.e. y=2x−32⋯(1) Given equation 4x+3y+9=0⋯(2) Solving both the equation, we get P≡(−920,−125)

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