Let P (2, -4) and Q (3, 1) be two given points. Let R (x, y) be a point such that ( x-2) (x-3) + (y-1) (y+4) = 0. If area of ΔPQR is 132 , then the number of possible positions of R are
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a
2
b
3
c
4
d
6
answer is A.
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Detailed Solution
ΔPQR is the right angle triangle radius = altitude