Question

If a point P is equidistant from A(1, 3), B(-3, 5) and C(5, -1), then PA is equal to

• (A) 5
• (B) 5√5
• (C) 25
• (D) 5√10

Answer 1

The Correct Option is D

Explanation:
Perpendicular bisector of $A(1,3)$ and $B(-3,5)$ is

$$\begin{array}{l}2x(x_1-x_2)+2y(y_1-y_2)=(x_1^2+y_1^2)\\ \Rightarrow 2x(1+3)+2y(3-5)=(1+9)\\ \Rightarrow -(x_2^2+y_2^2)\\ \Rightarrow 2x-4y=10-34\end{array}$$Similarly, perpendicular bisector of $A(1,3)$ and $C(5-1)$ is$$\begin{array}{c}2x(1-5)+2y(3+1)=(1+9)-(25+1)\\ -8x+8y=10-26\\ -x+y=-2\end{array}$$$$\Rightarrow x-y-2=0$$Point of intersection of Eqs. (i) and(ii), is $P=(-8,-10)$Then, the distance between $P$ and $A$ is$$\begin{array}{rl}PA& =\sqrt{(1+8{)}^2+(3+10{)}^2}\\ & =\sqrt{81+169}\\ & =\sqrt{250}=5\sqrt{10}\end{array}$$