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If the point $P (k – 1, 2)$ is equidistant from the points $A (3, k)$ and $B (k, 5)$, find the values of $k$.
Given:
The point $P (k-1, 2)$ is equidistant from the points $A(3, k)$ and $B(k, 5)$.
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,
\( \mathrm{PA}=\sqrt{(k-1-3)^{2}+(2-k)^{2}} \)
Squaring on both sides, we get,
\( \mathrm{PA}^{2}=(k-4)^{2}+(2-k)^{2} \)
\( =k^2-8k+16+4+k^{2}-4 k \)
\( =2k^{2}-12k+20 \)
\( \mathrm{PB}^{2}=(k-1-k)^{2}+(2-5)^{2} \)
\( =(-1)^{2}+(-3)^{2} \)
\( =1+9 \)
\( =10 \)
\( \Rightarrow 2k^{2}-12 k+20=10 \)
\( \Rightarrow 2(k^2-6k+10)=2(5) \)
\( \Rightarrow k^2-6k+10-5=0 \)
\( k^2-6k+5=0 \)
\( k^2-k-5k+5=0 \)
\( k(k-1)-5(k-1)=0 \)
\( (k-5)(k-1) =0 \)
\( k=5 \) or \( k=1 \)
The values of $k$ are $1$ and $5$. 
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