## problem statement

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

$A(0 ; 0 ; z)$

$B(3 ; 1 ; 3)$

$C(1 ; 4 ; 2)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(3-0)^{2}+(1-0)^{2}+(3-z)^{2}}=\sqrt{9+1+9-6 z+z^{2}}=\sqrt{z^{2}-6 z+19} \\
& A C=\sqrt{(1-0)^{2}+(4-0)^{2}+(2-z)^{2}}=\sqrt{1+16+4-4 z+z^{2}}=\sqrt{z^{2}-4 z+21}
\end{aligned}
$$

$$
\begin{aligned}
& \sqrt{z^{2}-6 z+19}=\sqrt{z^{2}-4 z+21} \\
& z^{2}-6 z+19=z^{2}-4 z+21 \\
& -2 z=2 \\
& z=-1
\end{aligned}
$$

Thus, $A(0 ; 0 ;-1)$.

## Problem Kuznetsov Analytic Geometry 11-3