## Problem Statement

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

$A(0 ; 0 ; z)$

$B(6 ;-7 ; 1)$

$C(-1 ; 2 ; 5)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(6-0)^{2}+(-7-0)^{2}+(1-z)^{2}}=\sqrt{36+49+1-2 z+z^{2}}=\sqrt{z^{2}-2 z+86} \\
& A C=\sqrt{(-1-0)^{2}+(2-0)^{2}+(5-z)^{2}}=\sqrt{1+4+25-10 z+z^{2}}=\sqrt{z^{2}-10 z+30}
\end{aligned}
$$

Since according to the problem $A B=A C$, then

$$
\sqrt{z^{2}-2 z+86}=\sqrt{z^{2}-10 z+30}
$$

$z^{2}-2 z+86=z^{2}-10 z+30$

$8 z=-56$

$z=-7$

Thus, $A(0 ; 0 ;-7)$.

## Problem Kuznetsov Analytic Geometry $11-10$