## Problem Statement

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

$A(0 ; y ; 0)$

$B(1 ; 6 ; 4)$

$C(5 ; 7 ; 1)$
## Solution

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

\[
\begin{aligned}
& A B=\sqrt{(1-0)^{2}+(6-y)^{2}+(4-0)^{2}}=\sqrt{1+36-12 y+y^{2}+16}=\sqrt{y^{2}-12 y+53} \\
& A C=\sqrt{(5-0)^{2}+(7-y)^{2}+(1-0)^{2}}=\sqrt{25+49-14 y+y^{2}+1}=\sqrt{y^{2}-14 y+75}
\end{aligned}
\]

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

\[
\sqrt{y^{2}-12 y+53}=\sqrt{y^{2}-14 y+75}
\]

\[
y^{2}-12 y+53=y^{2}-14 y+75
\]

\[
2 y=22
\]

\[
y=11
\]

Thus, $A(0 ; 11 ; 0)$.

## Problem Kuznetsov Analytic Geometry 11-13