## Problem Statement

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

$A(x ; 0 ; 0)$

$B(4 ; 6 ; 8)$

$C(2 ; 4 ; 6)$
## Solution

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

$A B=\sqrt{(4-x)^{2}+(6-0)^{2}+(8-0)^{2}}=\sqrt{16-8 x+x^{2}+36+64}=\sqrt{x^{2}-8 x+116}$

$A C=\sqrt{(2-x)^{2}+(4-0)^{2}+(6-0)^{2}}=\sqrt{4-4 x+x^{2}+16+36}=\sqrt{x^{2}-4 x+56}$

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

$$
\begin{aligned}
& \sqrt{x^{2}-8 x+116}=\sqrt{x^{2}-4 x+56} \\
& x^{2}-8 x+116=x^{2}-4 x+56 \\
& -4 x=-60 \\
& x=15
\end{aligned}
$$

Thus, $A(15 ; 0 ; 0)$.

## Problem Kuznetsov Analytic Geometry 11-29