A straight rod of length 4 units slides such that its ends $A$ and $B$ always lie on the $X$ and $Y$-axes…
A straight rod of length 4 units slides such that its ends $A$ and $B$ always lie on the $X$ and $Y$-axes respectively. Then, the locus of the centroid of $\triangle O A B$ is
$x^2+y^2=4$
$x^2+y^2=3$
$x^2+y^2=\frac{9}{16}$
$x^2+y^2=\frac{16}{9}$
Solution
Let the coordinate of vertices of $\triangle O A B$
$O=(0,0) A=(x, 0)$ and $B=(0, y)$
$(h, k)$ is centroid of $\triangle O A B$.
$\therefore \quad h=\frac{x}{3}$ and $k=\frac{y}{3}$
$\begin{aligned} & x=3 h \text { and } y=3 k \\ & x=3 h \text { and } y=3 k\end{aligned}$
Given, $x^2+y^2=16$
$x^2+y^2=9\left(h^2+k^2\right) \Rightarrow \frac{16}{9}=h^2+k^2$
$\therefore$ Locus of centroid of $\triangle O A B$ is $x^2+y^2=16 / 9$