Question

A variable plane is at a constant distance p from the origin cuts the coordinate axes in A, B and C. The locus of centroid of $△ABC$ is:

• A. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$
• B. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{3}{p^2}$
• C. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{9}{p^2}$
• D. None

Answer 1

Answer: (C)

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

$ax+by+cz=d$ be te variable plane.

distance from $(0,0,0)=\frac{d}{\sqrt{a^2+b^2+c^2}}=p$

Let $(h,x,l)$ be centroid .

$\therefore x,y,z$ intercepts are $\frac{d}{a},\frac{d}{b},\frac{d}{c}$

$\therefore G$ (centroid)$=(\frac{d}{3a},\frac{d}{3b},\frac{d}{3c})$

$h=\frac{d}{3a},k=\frac{d}{3b},l=\frac{d}{3b}\therefore \frac{1}{h^2}+\frac{1}{k^2}+\frac{1}{l^2}=\frac{9}{d^2}(a^2+b^2+c^2)=\frac{9}{p^2}$