Question

A point $P(2,-1)$ is equidistant from the points $(a,7)$ and $(-3,a)$. Find 'a'.

Answer 1

We know that the distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Let the given points be $A=(a,7)$ and $B=(-3,a)$ and the third point given is $P(2,-1)$.

We first find the distance between $P(2,-1)$ and $A=(a,7)$ as follows:

$PA=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(a-2)}^2+{(7-(-1\left)\right)}^2}=\sqrt{{(a-2)}^2+{(7+1)}^2}=\sqrt{{(a-2)}^2+8^2}$

$=\sqrt{{(a-2)}^2+64}$

Similarly, the distance between $P(2,-1)$ and $B=(-3,a)$ is:

$PB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(-3-2)}^2+{(a-(-1\left)\right)}^2}=\sqrt{{(-5)}^2+{(a+1)}^2}=\sqrt{25+{(a+1)}^2}$

Since the point $P(2,-1)$ is equidistant from the points $A(a,7)$ and $B=(-3,a)$, therefore, $PA=PB$ that is:

$\sqrt{{(a-2)}^2+64}=\sqrt{25+{(a+1)}^2}\Rightarrow (\sqrt{{(a-2)}^2+64}{)}^2=(\sqrt{25+{(a+1)}^2}{)}^2\Rightarrow {(a-2)}^2+64=25+{(a+1)}^2\Rightarrow {(a-2)}^2-{(a+1)}^2=25-64\Rightarrow {(a}^2+4-4a)-{(a}^2+1+2a)=-39(\because {(a-b)}^2=a^2+b^2-2ab,{(a+b)}^2=a^2+b^2+2ab)$

$\Rightarrow a^2+4-4a-a^2-1-2a=-39\Rightarrow -6a+3=-39\Rightarrow -6a=-39-3\Rightarrow -6a=-42\Rightarrow a=\frac{-42}{-6}=7$

Hence, $a=7$.