## Problem Statement

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

$A(0 ; y ; 0)$

$B(0 ; 5 ;-9)$

$C(-1 ; 0 ; 5)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(0-0)^{2}+(5-y)^{2}+(-9-0)^{2}}=\sqrt{25-10 y+y^{2}+81}=\sqrt{y^{2}-10 y+106} \\
& A C=\sqrt{(-1-0)^{2}+(0-y)^{2}+(5-0)^{2}}=\sqrt{1+y^{2}+25}=\sqrt{y^{2}+26}
\end{aligned}
$$

Since according to the problem $A B=A C$, then
$\sqrt{y^{2}-10 y+106}=\sqrt{y^{2}+26}$

$y^{2}-10 y+106=y^{2}+26$

$-10 y=-80$

$y=8$

Thus, $A(0 ; 8 ; 0)$.

## Problem Kuznetsov Analytical Geometry $11-18$