Question

The coordinates of the point equidistant from the four points O(0, 0, 0), A(a, 0, 0), B(0, b, 0) and C(0, 0, c) are ________________________.

Answer 1

Let us suppose any point be P(x, y, z) which is equidistant from O(0, 0, 0), A(a, 0, 0), B(0, b, 0) and C(0, 0, c)
$$\begin{gathered} \mathrm{i}.\mathrm{e}PO=\sqrt{x^2+y^2+z^2} \\ \therefore {\left(PO\right)}^2=x^2+y^2+z^2 \\ PA=\sqrt{{\left(x-a\right)}^2+y^2+z^2} \\ \mathrm{i}.\mathrm{e}{\left(PA\right)}^2={\left(x-a\right)}^2+y^2+z^2 \\ \mathrm{also}{\left(PB\right)}^2=x^2+{\left(y-b\right)}^2+c^2 \\ \mathrm{and}{\left(PC\right)}^2=x^2+y^2+{\left(z-c\right)}^2 \\ \mathrm{given}PO=PA=PB=PC \\ \mathrm{i}.\mathrm{e}{\left(PO\right)}^2={\left(PA\right)}^2={\left(PB\right)}^2={\left(PC\right)}^2 \\ \therefore \mathrm{We}\mathrm{get}, \\ {x}^2+y^2+z^2={\left(x-a\right)}^2+y^2+z^2 \\ {x}^2+y^2+z^2=x^2+{\left(y-b\right)}^2+z^2 \\ \mathrm{and}{x}^2+y^2+z^2=x^2+y^2+{\left(z-c\right)}^2 \\ \mathrm{i}.\mathrm{e} \\ {x}^2=x^2+a^2-2ax \\ {y}^2=y^2+b^2-2bx \\ \mathrm{and}{z}^2=z^2+c^2-2cx \\ \mathrm{i}.\mathrm{e}{a}^2=2ax\Rightarrow x=\frac{a}{2} \\ {b}^2=2by\Rightarrow y=\frac{b}{2} \\ \mathrm{and}{c}^2=2cz\Rightarrow z=\frac{c}{2} \end{gathered}$$
Hence, the coordinate of point equidistant from
O(0, 0, 0), A(a, 0, 0), B(0, b, 0) and C(0, 0, c) is $\left(\frac{a}{2},\frac{b}{2},\frac{c}{2}\right)$