## Problem Statement

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

$A(0 ; y ; 0)$

$B(-2 ; 4 ;-6)$

$C(8 ; 5 ; 1)$
## Solution

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

\[
\begin{aligned}
& A B=\sqrt{(-2-0)^{2}+(4-y)^{2}+(-6-0)^{2}}=\sqrt{4+16-8 y+y^{2}+36}=\sqrt{y^{2}-8 y+56} \\
& A C=\sqrt{(8-0)^{2}+(5-y)^{2}+(1-0)^{2}}=\sqrt{64+25-10 y+y^{2}+1}=\sqrt{y^{2}-10 y+90}
\end{aligned}
\]

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

\[
\sqrt{y^{2}-8 y+56}=\sqrt{y^{2}-10 y+90}
\]

\[
y^{2}-8 y+56=y^{2}-10 y+90
\]

\[
2 y=34
\]

\[
y=17
\]

Thus, $A(0 ; 17 ; 0)$.

## Problem Kuznetsov Analytic Geometry $11-19$