## Problem Statement

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

$A(0 ; 0 ; z)$

$B(-1 ;-1 ;-6)$

$C(2 ; 3 ; 5)$
## Solution

Let's find the distances $A B$ and $A C$:
$A B=\sqrt{(-1-0)^{2}+(-1-0)^{2}+(-6-z)^{2}}=\sqrt{1+1+36+12 z+z^{2}}=\sqrt{z^{2}+12 z+38}$ 
$A C=\sqrt{(2-0)^{2}+(3-0)^{2}+(5-z)^{2}}=\sqrt{4+9+25-10 z+z^{2}}=\sqrt{z^{2}-10 z+38}$

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

$$
\begin{aligned}
& \sqrt{z^{2}+12 z+38}=\sqrt{z^{2}-10 z+38} \\
& z^{2}+12 z+38=z^{2}-10 z+38 \\
& 22 z=0 \\
& z=0
\end{aligned}
$$

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

## Problem Kuznetsov Analytic Geometry 11-4