## Problem Statement

Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.

$A(0 ; 0 ; z)$

$B(7 ; 0 ;-15)$

$C(2 ; 10 ;-12)$
## Solution

Let's find the distances $A B$ and $A C$:

\[
\begin{aligned}
& A B=\sqrt{(7-0)^{2}+(0-0)^{2}+(-15-z)^{2}}=\sqrt{49+0+225+30 z+z^{2}}=\sqrt{z^{2}+30 z+274} \\
& A C=\sqrt{(2-0)^{2}+(10-0)^{2}+(-12-z)^{2}}=\sqrt{4+100+144+24 z+z^{2}}=\sqrt{z^{2}+24 z+248}
\end{aligned}
\]

Since by the condition of the problem $A B=A C$, then

\[
\sqrt{z^{2}+30 z+274}=\sqrt{z^{2}+24 z+248}
\]

\[
z^{2}+30 z+274=z^{2}+24 z+248
\]

\[
6 z=-26
\]

\[
z=-\frac{13}{3}=-4 \frac{1}{3}
\]

Thus, $A\left(0 ; 0 ;-4 \frac{1}{3}\right)$.

Problem Kuznetsov Analytical Geometry 11-11