## Problem Statement

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

$A(x ; 0 ; 0)$

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

$C(-1 ;-2 ;-3)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(-2-x)^{2}+(-4-0)^{2}+(-6-0)^{2}}=\sqrt{4+4 x+x^{2}+16+36}=\sqrt{x^{2}+4 x+56} \\
& A C=\sqrt{(-1-x)^{2}+(-2-0)^{2}+(-3-0)^{2}}=\sqrt{1+2 x+x^{2}+4+9}=\sqrt{x^{2}+2 x+14}
\end{aligned}
$$

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

$$
\begin{aligned}
& \sqrt{x^{2}+4 x+56}=\sqrt{x^{2}+2 x+14} \\
& x^{2}+4 x+56=x^{2}+2 x+14 \\
& 2 x=-42 \\
& x=-21
\end{aligned}
$$

Thus, $A(-21 ; 0 ; 0)$.

## Problem Kuznetsov Analytical Geometry 11-31