## Problem Statement

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

$A(x ; 0 ; 0)$

$B(1 ; 5 ; 9)$

$C(3 ; 7 ; 11)$
## Solution

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

$A B=\sqrt{(1-x)^{2}+(5-0)^{2}+(9-0)^{2}}=\sqrt{1-2 x+x^{2}+25+81}=\sqrt{x^{2}-2 x+107}$

$A C=\sqrt{(3-x)^{2}+(7-0)^{2}+(11-0)^{2}}=\sqrt{9-6 x+x^{2}+49+121}=\sqrt{x^{2}-6 x+179}$

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

$$
\sqrt{x^{2}-2 x+107}=\sqrt{x^{2}-6 x+179}
$$

$x^{2}-2 x+107=x^{2}-6 x+179$

$4 x=72$

$x=18$

Thus, $A(18 ; 0 ; 0)$.

## Problem Kuznetsov Analytical Geometry 11-28