## Problem Statement

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

$A(0 ; y ; 0)$

$B(2 ; 2 ; 4)$

$C(0 ; 4 ; 2)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(2-0)^{2}+(2-y)^{2}+(4-0)^{2}}=\sqrt{4+4-4 y+y^{2}+16}=\sqrt{y^{2}-4 y+24} \\
& A C=\sqrt{(0-0)^{2}+(4-y)^{2}+(2-0)^{2}}=\sqrt{0+16-8 y+y^{2}+4}=\sqrt{y^{2}-8 y+20}
\end{aligned}
$$

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

$$
\begin{aligned}
& \sqrt{y^{2}-4 y+24}=\sqrt{y^{2}-8 y+20} \\
& y^{2}-4 y+24=y^{2}-8 y+20 \\
& 4 y=-4 \\
& y=-1
\end{aligned}
$$

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

## Problem Kuznetsov Analytic Geometry 11-16