Question

A variable plane which remains at a constant distance $3p$ from the origin cuts the coordinate axes at $A,B,C$. Show that the locus of the centroid of triangle $ABC$ is $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$

Answer 1

Let the equation of plane be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ , where $a,b,c$ are intercepts of plane on $x,y,z$ axis respectively.

The distance from origin to plane will be $\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}=3p$

$\Rightarrow \frac{1}{a^2}+\frac{1}{c^2}+\frac{1}{b^2}=\frac{1}{9p^2}$

The centroid of triangle $ABC$ is $(\frac{a}{3},\frac{b}{3},\frac{c}{3})$ and let it be $(x,y,z)$

So we have $a=3x,b=3y,c=3z$

$\Rightarrow \frac{1}{9x^2}+\frac{1}{9y^2}+\frac{1}{9z^2}=\frac{1}{9p^2}$

$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$

Hence proved