Coordinate Geometry Question 157
The point P is equidistant from A(1,3), B (-3,5) and C(5,-1). Then PA =
[EAMCET 2003]
Options:
5
B) $ 5\sqrt{5} $
25
D) $ 5\sqrt{10} $
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Answer:
Correct Answer: D
Solution:
Perpendicular bisector of $ A(1,3) $ and $ B(-3,5) $ is $ 2x(1+3)+2y(3-5)=(1^{2}+3^{2})-((-3)^{2}+5^{2}) $
$ \Rightarrow 2x(1+3)+2y(3-5)=(1+9)-(9+25) $
$ \Rightarrow 2x-y+6=0 $ ……(i) Perpendicular bisector of $ A(1,3) $ and $ C(5,-1) $ is $ 2x(5-1)+2y(-1-3)=(5+1)-(1+9) $
$ \Rightarrow x-y-2=0 $
…..(ii) Point of intersection of (i) and (ii) is $ P=(-8,-10) $ Then $ PA=\sqrt{{{(1+8)}^{2}}+{{(3+10)}^{2}}}=\sqrt{81+169} $
$ =\sqrt{250}=5\sqrt{10} $ .
BETA
