## Problem Statement

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

$A(0 ; 0 ; z)$

$B(-5 ;-5 ; 6)$

$C(-7 ; 6 ; 2)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(-5-0)^{2}+(-5-0)^{2}+(6-z)^{2}}=\sqrt{25+25+36-12 z+z^{2}}=\sqrt{z^{2}-12 z+86} \\
& A C=\sqrt{(-7-0)^{2}+(6-0)^{2}+(2-z)^{2}}=\sqrt{49+36+4-4 z+z^{2}}=\sqrt{z^{2}-4 z+89}
\end{aligned}
$$

Since by the condition of the problem $A B=A C$, then
$\sqrt{z^{2}-12 z+86}=\sqrt{z^{2}-4 z+89}$

$z^{2}-12 z+86=z^{2}-4 z+89$

$-8 z=3$

$z=-\frac{3}{8}$

Thus, $A\left(0 ; 0 ;-\frac{3}{8}\right)$.

Problem Kuznetsov Analytic Geometry 11-6