## Problem Statement

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

$A(x ; 0 ; 0)$

$B(8 ; 1 ;-7)$

$C(10 ;-2 ; 1)$
## Solution

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

$A B=\sqrt{(8-x)^{2}+(1-0)^{2}+(-7-0)^{2}}=\sqrt{64-16 x+x^{2}+1+49}=\sqrt{x^{2}-16 x+114}$

$A C=\sqrt{(10-x)^{2}+(-2-0)^{2}+(1-0)^{2}}=\sqrt{100-20 x+x^{2}+4+1}=\sqrt{x^{2}-20 x+105}$

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

$$
\begin{aligned}
& \sqrt{x^{2}-16 x+114}=\sqrt{x^{2}-20 x+105} \\
& x^{2}-16 x+114=x^{2}-20 x+105 \\
& 4 x=-9 \\
& x=-\frac{9}{4}=-2.25
\end{aligned}
$$

Thus, $A(-2.25 ; 0 ; 0)$.

## Problem Kuznetsov Analytical Geometry 11-24