## Problem Statement

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

$A(x ; 0 ; 0)$

$B(1 ; 2 ; 3)$

$C(2 ; 6 ; 10)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(1-x)^{2}+(2-0)^{2}+(3-0)^{2}}=\sqrt{1-2 x+x^{2}+4+9}=\sqrt{x^{2}-2 x+14} \\
& A C=\sqrt{(2-x)^{2}+(6-0)^{2}+(10-0)^{2}}=\sqrt{4-4 x+x^{2}+36+100}=\sqrt{x^{2}-4 x+140}
\end{aligned}
$$

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

$$
\begin{aligned}
& \sqrt{x^{2}-2 x+14}=\sqrt{x^{2}-4 x+140} \\
& x^{2}-2 x+14=x^{2}-4 x+140 \\
& 2 x=126 \\
& x=63
\end{aligned}
$$

Thus, $A(63 ; 0 ; 0)$.

## Problem Kuznetsov Analytical Geometry $11-30$