## Problem Statement

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

$A(0 ; 0 ; z)$

$B(-13 ; 4 ; 6)$

$C(10 ;-9 ; 5)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(-13-0)^{2}+(4-0)^{2}+(6-z)^{2}}=\sqrt{169+16+36-12 z+z^{2}}=\sqrt{z^{2}-12 z+221} \\
& A C=\sqrt{(10-0)^{2}+(-9-0)^{2}+(5-z)^{2}}=\sqrt{100+81+25-10 z+z^{2}}=\sqrt{z^{2}-10 z+206}
\end{aligned}
$$

Since by the condition of the problem $A B=A C$, then
$\sqrt{z^{2}-12 z+221}=\sqrt{z^{2}-10 z+206}$

$z^{2}-12 z+221=z^{2}-10 z+206$

$-2 z=-15$

$z=\frac{15}{2}=7.5$

Thus, $A(0 ; 0 ; 7.5)$.

## Problem Kuznetsov Analytical Geometry $11-5$