# 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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Updated on: 10-Oct-2022

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