## Problem Statement

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

$A(0 ; y ; 0)$

$B(0 ;-4 ; 1)$

$C(1 ;-3 ; 5)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(0-0)^{2}+(-4-y)^{2}+(1-0)^{2}}=\sqrt{0+16+8 y+y^{2}+1}=\sqrt{y^{2}+8 y+17} \\
& A C=\sqrt{(1-0)^{2}+(-3-y)^{2}+(5-0)^{2}}=\sqrt{1+9+6 y+y^{2}+25}=\sqrt{y^{2}+6 y+35}
\end{aligned}
$$

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

$\sqrt{y^{2}+8 y+17}=\sqrt{y^{2}+6 y+35}$
$y^{2}+8 y+17=y^{2}+6 y+35$

$2 y=18$

$y=9$

Thus, $A(0 ; 9 ; 0)$.

## Problem Kuznetsov Analytic Geometry 11-17