Question

Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Answer 1

Let P (x, y, z) be any point that is equidistant from the points A (1, 2, 3) and B (3, 2, −1).
Then, we have:
PA = PB
$$\begin{gathered} \Rightarrow \sqrt{{\left(x-1\right)}^2+{\left(y-2\right)}^2+{\left(z-3\right)}^2}=\sqrt{{\left(x-3\right)}^2+{\left(y-2\right)}^2+{\left(z+1\right)}^2} \\ \Rightarrow x^2-2x-1+y^2-4y+4+z^2-6z+9=x^2-6x+9+y^2-4y+4+z^2+2z+1 \\ \Rightarrow -2x-4y-6z+14=-6x-4y+2z+14 \\ \Rightarrow -2x+6x-4y+4y-6z-2z=14-14 \\ \Rightarrow 4x-8z=0 \\ \Rightarrow x-2z=0 \end{gathered}$$
Hence, the locus is x $-$ 2z = 0