Question

Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).

Answer 1

Let the point on the y-axis be Y $\left(0,y,0\right)$ which is equidistant from the points P $\left(3,1,2\right)$ and Q $\left(5,5,2\right)$.
Then ,PY=QY
Now, $\sqrt{{\left(0-3\right)}^2+{\left(y-1\right)}^2+{\left(0-2\right)}^2}$=$\sqrt{{\left(0-5\right)}^2+{\left(y-5\right)}^2+{\left(0-2\right)}^2}$

$$\begin{gathered} \Rightarrow \sqrt{{\left(-3\right)}^2+y^2+2y+1+{\left(-2\right)}^2}=\sqrt{{\left(-5\right)}^2+{\left(y-5\right)}^2+{\left(-2\right)}^2} \\ \Rightarrow \sqrt{9+y^2+2y+1+4}=\sqrt{25+y^2+10y+25+4} \\ \Rightarrow \sqrt{y^2+2y+14}=\sqrt{y^2+10y+54} \end{gathered}$$
$\Rightarrow y^2+2y+14=y^2+10y+54$

$$\begin{gathered} 2y-10y=54-14 \\ -8y=40 \\ -y=\frac{40}{8} \\ y=-5 \end{gathered}$$

Thus, the required point on the y-axis is (0, −5, 0).