Question

The points $P$ is equidistant from $A(1,3)$, $B(-3,5)$ and $C(5,-1)$, then $PA$ is equal to

• A. $5$
• B. $5\sqrt{5}$
• C. $25$
• D. $5\sqrt{10}$

Answer 1

Answer: (D)

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

$2x\left(x_1-x_2\right)+2y\left(y_1-y_2\right)=\left(x_1^2+y_1^2\right)-\left(x_2^2+y_2^2\right)$
$\Rightarrow 2x(1+3)+2y(3-5)=(1+9)-(9+25)$
$\Rightarrow 8x-4y=10-34$
$\Rightarrow 2x-y+6=0\dots (i)$
Similarly, perpendicular bisector of $A(1,3)$ and $C(5,-1)$ is
$2x(1-5)+2y(3+1)=(1+9)-(25+1)$
$\Rightarrow -8x+8y=10-26$
$\Rightarrow -x+y=-2$
$\Rightarrow x-y-2=0\dots (ii)$
Point of intersection of Eqs. $(i)$ and $(ii)$, is
$P=(-8,-10)$.
Then, the distance between $P$ and $A$ is
$PA=\sqrt{(1+8{)}^2+(3+10{)}^2}$
$=\sqrt{81+169}$
$=\sqrt{250}=5\sqrt{10}$