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 given variable plane meets X, Y and Z axes at A(a, 0, 0), B(0, b, 0), C(0, 0, c).

Therefore the equation of given plane is given by

Let \((\alpha, \beta, \gamma)\) be the coordinates of the centroid of triangle ABC. Then

\(\because\) 3p is the distance from origin to the plane (i)

Squaring both sides, we have

Therefore, Locus of \((\alpha, \beta, \gamma)\) is \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}\) Hence proved.