## Problem Statement

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

$A(x ; 0 ; 0)$

$B(-2 ; 0 ; 6)$

$C(0 ;-2 ;-4)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(-2-x)^{2}+(0-0)^{2}+(6-0)^{2}}=\sqrt{4+4 x+x^{2}+0+36}=\sqrt{x^{2}+4 x+40} \\
& A C=\sqrt{(0-x)^{2}+(-2-0)^{2}+(-4-0)^{2}}=\sqrt{x^{2}+4+16}=\sqrt{x^{2}+20}
\end{aligned}
$$

Since according to the problem's condition $A B=A C$, then

$$
\begin{aligned}
& \sqrt{x^{2}+4 x+40}=\sqrt{x^{2}+20} \\
& x^{2}+4 x+40=x^{2}+20 \\
& 4 x=-20 \\
& x=-5
\end{aligned}
$$

Thus, $A(-5 ; 0 ; 0)$.

## Problem Kuznetsov Analytic Geometry 11-27