## Problem Statement

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

$A(x ; 0 ; 0)$

$B(3 ; 5 ; 6)$

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

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

$$
\begin{aligned}
& A B=\sqrt{(3-x)^{2}+(5-0)^{2}+(6-0)^{2}}=\sqrt{9-6 x+x^{2}+25+36}=\sqrt{x^{2}-6 x+70} \\
& A C=\sqrt{(1-x)^{2}+(2-0)^{2}+(3-0)^{2}}=\sqrt{1-2 x+x^{2}+4+9}=\sqrt{x^{2}-2 x+14}
\end{aligned}
$$

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

$$
\sqrt{x^{2}-6 x+70}=\sqrt{x^{2}-2 x+14}
$$

$x^{2}-6 x+70=x^{2}-2 x+14$

$-4 x=-56$

$x=14$

Thus, $A(14 ; 0 ; 0)$.

Problem Kuznetsov Analytical Geometry 11-25