## Problem Statement

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

$A(0 ; 0 ; z)$

$B(10 ; 0 ;-2)$

$C(9 ;-2 ; 1)$
## Solution

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

$A B=\sqrt{(10-0)^{2}+(0-0)^{2}+(-2-z)^{2}}=\sqrt{100+0+4+4 z+z^{2}}=\sqrt{z^{2}+4 z+104}$

$A C=\sqrt{(9-0)^{2}+(-2-0)^{2}+(1-z)^{2}}=\sqrt{81+4+1-2 z+z^{2}}=\sqrt{z^{2}-2 z+86}$

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

$$
\begin{aligned}
& \sqrt{z^{2}+4 z+104}=\sqrt{z^{2}-2 z+86} \\
& z^{2}+4 z+104=z^{2}-2 z+86 \\
& 6 z=-18 \\
& z=-3
\end{aligned}
$$

Thus, $A(0 ; 0 ;-3)$.

## Problem Kuznetsov Analytic Geometry 11-8