## Problem Statement

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

$A(0 ; 0 ; z)$

$B(5 ; 1 ; 0)$

$C(0 ; 2 ; 3)$
## Solution

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

$A B=\sqrt{(5-0)^{2}+(1-0)^{2}+(0-z)^{2}}=\sqrt{25+1+z^{2}}=\sqrt{z^{2}+26}$

$A C=\sqrt{(0-0)^{2}+(2-0)^{2}+(3-z)^{2}}=\sqrt{0+4+9-6 z+z^{2}}=\sqrt{z^{2}-6 z+13}$

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

$\sqrt{z^{2}+26}=\sqrt{z^{2}-6 z+13}$

$z^{2}+26=z^{2}-6 z+13$

$6 z=-13$

$z=-\frac{13}{6}=-2 \frac{1}{6}$

Thus, $A\left(0 ; 0 ;-2 \frac{1}{6}\right)$.

## Problem Kuznetsov Analytical Geometry 11-1