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If the point $P (2, 2)$ is equidistant from the points $A (-2, k)$ and $B (-2k, -3)$, find $k$. Also, find the length of AP.
Given:
The point $P (2, 2)$ is equidistant from the points $A (-2, k)$ and $B (-2k, -3)$.
To do:
We have to find the value of $k$ and the length of AP.
Solution:
Point $P (2, 2)$ is equidistant from the points $A (-2, k)$ and $B (-2k, -3)$.
This implies,
$PA = PB$
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,
\( \sqrt{(2+2)^{2}+(2-k)^{2}}=\sqrt{(2+2 k)^{2}+(2+3)^{2}} \)
\( (2+2)^{2}+(2-k)^{2}=(2+2 k)^{2}+(2+3)^{2} \)
\( (4)^{2}+(2-k)^{2}=(2+2 k)^{2}+(5)^{2} \)
\( 16+4+k^{2}-4 k=4+4 k^{2}+8 k+25 \)
\( 4 k^{2}+8 k+29-16-4-k^{2}+4 k=0 \)
\( 3 k^{2}+12 k+9=0 \)
\( k^{2}+4 k+3=0 \)
\( k^{2}+k+3 k+3=0 \)
\( k(k+1)+3(k+1)=0 \)
\( (k+1)(k+3)=0 \)
\( k+1=0 \) or \( k+3=0 \)
\( k=-1 \) or \( k=-3 \)
\( \therefore k=-1,-3 \)
\( \mathrm{AP}=\sqrt{(4)^{2}+(2-k)^{2}} \)
\( =\sqrt{16+(2+1)^{2}} \)
\( =\sqrt{16+9} \)
\( =\sqrt{25} \)
\( =5 \)
Therefore, the values of $k$ are $-1$ and $-3$ and the distance $AP$ is $5$.
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