A variable line passing through a fixed point $\left(\mathrm{\alpha },\mathrm{\beta }\right)$ intersects the coordinate axes at A and B.If O is the origin, then the locus of the centroid of the triangle OAB is

let $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$ be the line which cuts coordinate axes at A and B

i.e A=(a, 0), B=(0, b)

Since it passes through $\left(\mathrm{\alpha }, \mathrm{\beta }\right)$ then 

$\frac{\mathrm{\alpha }}{\mathrm{a}}+\frac{\mathrm{\beta }}{\mathrm{b}}=1 - \left(1\right)$

Let P(x, y) be locus of centriod of ${∆}^{\mathrm{le}} \mathrm{OAB}$

$$\begin{gathered} \Rightarrow (\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}\right) \\ \Rightarrow \mathrm{x}=\frac{\mathrm{a}}{3}, \mathrm{y}=\frac{\mathrm{b}}{3} \\ \Rightarrow \mathrm{a}=3\mathrm{x}, \mathrm{b}=3\mathrm{y} \\ \mathrm{from} \left(1\right) \\ \frac{\mathrm{\alpha }}{3\mathrm{x}}+\frac{\mathrm{\beta }}{3\mathrm{y}}=1 \\ \Rightarrow \frac{\mathrm{\alpha y}+\mathrm{\beta x}}{3\mathrm{xy}}=1 \\ \Rightarrow \mathrm{\alpha y}+\mathrm{\beta x}=3\mathrm{xy} \\ \Rightarrow \mathrm{\alpha y}+\mathrm{\beta x}-3\mathrm{xy}=0 \end{gathered}$$