## Problem Statement

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

$A(0 ; y ; 0)$

$B(7 ; 3 ;-4)$

$C(1 ; 5 ; 7)$
## Solution

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

$$
\begin{aligned}
& A B=\sqrt{(7-0)^{2}+(3-y)^{2}+(-4-0)^{2}}=\sqrt{49+9-6 y+y^{2}+16}=\sqrt{y^{2}-6 y+74} \\
& A C=\sqrt{(1-0)^{2}+(5-y)^{2}+(7-0)^{2}}=\sqrt{1+25-10 y+y^{2}+49}=\sqrt{y^{2}-10 y+75}
\end{aligned}
$$

Since by the condition of the problem $A B=A C$, then
$\sqrt{y^{2}-6 y+74}=\sqrt{y^{2}-10 y+75}$

$y^{2}-6 y+74=y^{2}-10 y+75$

$4 y=1$

$y=\frac{1}{4}$

Thus, $A\left(0 ; \frac{1}{4} ; 0\right)$.

## Problem Kuznetsov Analytical Geometry $11-20$