## problem statement

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

$A(0 ; y ; 0)$

$B(3 ; 0 ; 3)$

$C(0 ; 2 ; 4)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(3-0)^{2}+(0-y)^{2}+(3-0)^{2}}=\sqrt{9+y^{2}+9}=\sqrt{y^{2}+18} \\
& A C=\sqrt{(0-0)^{2}+(2-y)^{2}+(4-0)^{2}}=\sqrt{4-4 y+y^{2}+16}=\sqrt{y^{2}-4 y+20}
\end{aligned}
$$

Since according to the problem $A B=A C$, then

$\sqrt{y^{2}+18}=\sqrt{y^{2}-4 y+20}$

$y^{2}+18=y^{2}-4 y+20$

$4 y=2$

$y=\frac{1}{2}$

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

## Problem Kuznetsov Analytic Geometry 11-12