## Problem Statement

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

$A(0 ; y ; 0)$

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

$C(7 ; 2 ; 5)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(-2-0)^{2}+(-4-y)^{2}+(6-0)^{2}}=\sqrt{4+16+8 y+y^{2}+36}=\sqrt{y^{2}+8 y+56} \\
& A C=\sqrt{(7-0)^{2}+(2-y)^{2}+(5-0)^{2}}=\sqrt{49+4-4 y+y^{2}+25}=\sqrt{y^{2}-4 y+78}
\end{aligned}
$$

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

$\sqrt{y^{2}+8 y+56}=\sqrt{y^{2}-4 y+78}$
$y^{2}+8 y+56=y^{2}-4 y+78$

$12 y=22$

$y=\frac{11}{6}=1 \frac{5}{6}$

Thus, $A\left(0 ; 1 \frac{5}{6} ; 0\right)$.

## Problem Kuznetsov Analytic Geometry $11-15$