## Problem Statement

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

$A(0 ; 0 ; z)$

$B(-6 ; 7 ; 5)$

$C(8 ;-4 ; 3)$
## Solution

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

$A B=\sqrt{(-6-0)^{2}+(7-0)^{2}+(5-z)^{2}}=\sqrt{36+49+25-10 z+z^{2}}=\sqrt{z^{2}-10 z+110}$

$A C=\sqrt{(8-0)^{2}+(-4-0)^{2}+(3-z)^{2}}=\sqrt{64+16+9-6 z+z^{2}}=\sqrt{z^{2}-6 z+89}$

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

$$
\begin{aligned}
& \sqrt{z^{2}-10 z+110}=\sqrt{z^{2}-6 z+89} \\
& z^{2}-10 z+110=z^{2}-6 z+89
\end{aligned}
$$

$-4 z=-21$

$z=5.25$

Thus, $A(0 ; 0 ; 5.25)$.

## Problem Kuznetsov Analytical Geometry $11-9$