Question

The point $\mathrm{P}$ is equidistant from $\mathrm{A}(1,3),\mathrm{B}(-3,5)$ and $\mathrm{C}(5,-1)$, then $\mathrm{PA}$ is equal to

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

Answer 1

The correct option is D

$5\sqrt{10}$

Compute the required value:

Let the coordinates of point $\mathrm{P}$ is $\left(\mathrm{x},\mathrm{y}\right)$.

Given: Point $\mathrm{P}$ is equidistant from $\mathrm{A}(1,3),\mathrm{B}(-3,5)$ and $\mathrm{C}(5,-1)$.

The perpendicular bisector of $\mathrm{AB}$ is given by,

$2\mathrm{x}({\mathrm{x}}_1-{\mathrm{x}}_2)+2\mathrm{y}({\mathrm{y}}_1-{\mathrm{y}}_2)=\left({{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2\right)-\left({{\mathrm{x}}_2}^2+{{\mathrm{y}}_2}^2\right)$

$\Rightarrow$$2x(1+3)+2y\left(3-5\right)=(1+9)-\left(9+25\right)$

$\Rightarrow$ $8x-4y=-24$

$\Rightarrow$ $2x-y=-6$……………..$\left(1\right)$

The perpendicular bisector of $\mathrm{AC}$ is given by,

$2\mathrm{x}({\mathrm{x}}_1-{\mathrm{x}}_2)+2\mathrm{y}({\mathrm{y}}_1-{\mathrm{y}}_2)=\left({{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2\right)-\left({{\mathrm{x}}_2}^2+{{\mathrm{y}}_2}^2\right)$

$\Rightarrow$ $2x\left(1-5\right)+2y(3+1)=(1+9)-(25+1)$

$\Rightarrow$ $-8x+8y=-16$

$\Rightarrow$ $x-y=2$………..….$\left(2\right)$

Subtract equation $\left(1\right)$ and $\left(2\right)$,

$2x-y-x+y=-6-2$

$\Rightarrow$ $x=-8$

Substitute the value of $x$ in equation $\left(1\right)$,

$$\begin{gathered} 2(-8)-y=-6 \\ \Rightarrow y=-10 \end{gathered}$$

Therefore, the coordinates of point $\mathrm{P}$ is $\left(-8,-10\right)$.

Hence, $\mathrm{PA}=\sqrt{{\left(1+8\right)}^2+{\left(3+10\right)}^2}$

$\Rightarrow$ $\mathrm{PA}=\sqrt{81+169}$

$\Rightarrow$ $\mathrm{PA}=\sqrt{250}$

$\Rightarrow$ $\mathrm{PA}=5\sqrt{10}$

Hence option$\left(D\right)$ is the correct option.