## Problem Statement

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

$A(x ; 0 ; 0)$

$B(4 ; 5 ;-2)$

$C(2 ; 3 ; 4)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(4-x)^{2}+(5-0)^{2}+(-2-0)^{2}}=\sqrt{16-8 x+x^{2}+25+4}=\sqrt{x^{2}-8 x+45} \\
& A C=\sqrt{(2-x)^{2}+(3-0)^{2}+(4-0)^{2}}=\sqrt{4-4 x+x^{2}+9+16}=\sqrt{x^{2}-4 x+29}
\end{aligned}
$$

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

$$
\begin{aligned}
& \sqrt{x^{2}-8 x+45}=\sqrt{x^{2}-4 x+29} \\
& x^{2}-8 x+45=x^{2}-4 x+29 \\
& -4 x=-16 \\
& x=4
\end{aligned}
$$

Thus, $A(4 ; 0 ; 0)$

## Problem Kuznetsov Analytic Geometry 11-26