If the first point of trisection of $\mathrm{AB}$ is $\left(\mathrm{t}, 2\mathrm{t}\right)$ and the ends $\mathrm{A}, \mathrm{B}$ moves on $\mathrm{x}$ and $\mathrm{y}$ axis respectively, then locus of midpoint of $\mathrm{AB}$ is

let $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$ be line which Cuts x-axis at A=(a, 0), y-axis at B=(0, b)

Let P=(x, y) be locus of mid point of $\overline{\mathrm{AB}}$

$$\begin{gathered} \Rightarrow (\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right) \\ \Rightarrow \mathrm{a}=2\mathrm{x}, \mathrm{b}=2\mathrm{y} -\left(1\right) \end{gathered}$$

Given that (t, 2t) is first point of trisection of  $\overline{\mathrm{AB}}$

$$\begin{gathered} \Rightarrow (\mathrm{t}, 2\mathrm{t})=\left(\frac{0+2\mathrm{a}}{3}, \frac{\mathrm{b}+0}{3}\right) \\ \Rightarrow \mathrm{t}=\frac{2\mathrm{a}}{3}, 2\mathrm{t}=\frac{\mathrm{b}}{3} \\ \Rightarrow 2\mathrm{a}=3\mathrm{t}, \mathrm{t}=\frac{\mathrm{b}}{6} \\ \Rightarrow 2\mathrm{a}=3\left(\frac{\mathrm{b}}{6}\right) \\ \Rightarrow 2\left(2\mathrm{x}\right)=\frac{1}{2}\left(2\mathrm{y}\right) \left(\because \mathrm{from} \left(1\right)\right) \\ \Rightarrow 4\mathrm{x}=\mathrm{y} \end{gathered}$$