The point on the line $4 x-y-2=0$ which is equidistant from the points $(-5,6)$ and $(3,2)$ is

Question

Aicharya · Accepted Answer

Let the required point be $\mathrm{P}\left(x_1, y_1\right)$ and it is on the line $4 x-y-2=0$. Then,

Also given, point $P$ is equidistant from $A(-5,6)$ and $\begin{aligned} & B(3,2) . \\ & \therefore \quad P A^2=P B^2 \\ & \left(x_1+5\right)^2+\left(y_1-6\right)^2=\left(x_1-3\right)^2+\left(y_1-2\right)^2 \\ & x_1^2+10 x_1+25+y_1^2-12 y_1+36 \\ & =x_1^2-6 x_1+9+y_1^2-4 y_1+4 \\ & =16 x_1-8 y_1+48=0 \end{aligned}$

On solving Eq. (i) in Eq. (ii), we get $\begin{aligned} & \therefore \quad x_1=4 \\ & \text { and } y_1=14 \end{aligned}$

The point on the line $4 x-y-2=0$ which is equidistant from the points $(-5,6)$ and $(3,2)$ is

Solution