Coordinate Geometry Question 157
Question: The points P is equidistant from A(1,3), B (-3,5) and C(5,-1). Then PA =
[EAMCET 2003]
Options:
A) 5
B) $ 5\sqrt{5} $
C) 25
D) $ 5\sqrt{10} $
Show Answer
Answer:
Correct Answer: D
Solution:
Perpendicular bisector of $ A(1,3) $ and $ B(-3,5) $ is $ 2x(x_1-x_2)+2y(y_1-y_2)=(x_1^{2}+y_1^{2})-(x_2^{2}+y_2^{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(1-5)+2y(3+1)=(1+9)-(25+1) $
$ \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
