## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(0-0)^{2}+(-2-y)^{2}+(4-0)^{2}}=\sqrt{0+4+4 y+y^{2}+16}=\sqrt{y^{2}+4 y+20} \\
& A C=\sqrt{(-4-0)^{2}+(0-y)^{2}+(4-0)^{2}}=\sqrt{16+y^{2}+16}=\sqrt{y^{2}+32}
\end{aligned}
$$

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

$\sqrt{y^{2}+4 y+20}=\sqrt{y^{2}+32}$

$y^{2}+4 y+20=y^{2}+32$

$4 y=12$

$y=3$

Thus, $A(0 ; 3 ; 0)$.

## Problem Kuznetsov Analytical Geometry 11-21
## Problem Statement

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

$A(0 ; y ; 0)$

$B(0 ;-2 ; 4)$

$C(-4 ; 0 ; 4)$