Find the value of $k$, if the point $P (0, 2)$ is equidistant from $(3, k)$ and $(k, 5)$.

Given:

The point $P (0, 2)$ is equidistant from $(3, k)$ and $(k, 5)$.

To do:

We have to find the value of $k$.

Solution:

$PA$ is equidistant from $PB$.

This implies,

$PA=PB$

Squaring on both sides, we get,

$PA^2=PB^2$

We know that,

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

Therefore,

\( \mathrm{PA}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)

\( =\sqrt{(0-3)^{2}+(2-k)^{2}} \)

\( \mathrm{PA}^{2}=(0-3)^{2}+(2-k)^{2} \)

\( =(-3)^{2}+(2-k)^{2} \)

\( =9+4+k^{2}-4 k \)

\( =k^{2}-4 k+13 \) 

\( \mathrm{PB}^{2}=(k-0)^{2}+(5-2)^{2} \)

\( =k^{2}+(3)^{2} \)

\( =k^{2}+9 \)

\( \Rightarrow k^{2}-4 k+13=k^{2}+9 \)

\( \Rightarrow -4 k=9-13 \)

\( \Rightarrow-4 k=-4 \)

\( k=\frac{-4}{-4}=1 \)

The value of $k$ is $1$.