## Problem Statement

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

$A(0 ; 0 ; z)$

$B(3 ; 3 ; 1)$

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

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

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

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

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

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

## Problem Kuznetsov Analytic Geometry 11-2