## Problem Statement

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

$A(0 ; y ; 0)$

$B(-2 ; 8 ; 10)$

$C(6 ; 11 ;-2)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(-2-0)^{2}+(8-y)^{2}+(10-0)^{2}}=\sqrt{4+64-16 y+y^{2}+100}=\sqrt{y^{2}-16 y+168} \\
& A C=\sqrt{(6-0)^{2}+(11-y)^{2}+(-2-0)^{2}}=\sqrt{36+121-22 y+y^{2}+4}=\sqrt{y^{2}-22 y+161}
\end{aligned}
$$

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

$$
\begin{aligned}
& \sqrt{y^{2}-16 y+168}=\sqrt{y^{2}-22 y+161} \\
& y^{2}-16 y+168=y^{2}-22 y+161 \\
& 6 y=-7 \\
& y=-\frac{7}{6}
\end{aligned}
$$

$$
\text { Thus } A\left(0 ;-\frac{7}{6} ; 0\right) .
$$

## Problem Kuznetsov Analytic Geometry 11-14