A straight rod of length 3l unit slides with its ends A,B alwalys on the x and y axes respectively. Then the locus of the centroid of $∆O\mathrm{AB}$ is

let $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$ be rod cuts co-ordinate axes at A & B

Where A = (a, 0), B = (0, b) and O=(0,0)

Given AB=3l 

$$\begin{gathered} \Rightarrow {\mathrm{AB}}^2=9{\mathrm{l}}^2 \\ \Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2=9{\mathrm{l}}^2 - \left(1\right) \\ \mathrm{let} ({\mathrm{x}}_1, {\mathrm{y}}_1) \mathrm{be} \mathrm{locus} \mathrm{of} \mathrm{centriod} \mathrm{of} {∆}^{\mathrm{le}} O\mathrm{AB} \\ ({\mathrm{x}}_1, {\mathrm{y}}_1)=\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}\right) \\ \Rightarrow {\mathrm{x}}_1=\frac{\mathrm{a}}{3}, {\mathrm{y}}_1=\frac{\mathrm{b}}{3} \\ \Rightarrow \mathrm{a}=3{\mathrm{x}}_1, \mathrm{b}=3{\mathrm{y}}_1 \\ \mathrm{from} \left(1\right) \\ 9{{\mathrm{x}}_1}^2+9{{\mathrm{y}}_1}^2=9{\mathrm{l}}^2 \\ \therefore \mathrm{locus} \mathrm{of} \mathrm{P} \mathrm{is} {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{l}}^2 \end{gathered}$$