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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 $1$.
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