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(D) Let the coordinates of the vertices of $	riangle OAB$ be $O(0,0)$,$A(a, 0)$,and $B(0, b)$.Since the length of the rod $AB$ is $4$,we have $a^2 +

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$x^2+y^2=\frac{16}{9}$

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(D) Let the coordinates of the vertices of $	riangle OAB$ be $O(0,0)$,$A(a, 0)$,and $B(0, b)$.Since the length of the rod $AB$ is $4$,we have $a^2 + b^2 = 4^2 = 16$.Let $(x, y)$ be the centroid of $	riangle OAB$.Then,$x = \frac{0+a+0}{3} = \frac{a}{3} \implies a = 3x$.And $y = \frac{0+0+b}{3} = \frac{b}{3} \implies b = 3y$.Substituting these into the equation $a^2 + b^2 = 16$,we get $(3x)^2 + (3y)^2 = 16$.$9x^2 + 9y^2 = 16$.$x^2 + y^2 = \frac{16}{9}$.Thus,the locus of the centroid is $x^2 + y^2 = \frac{16}{9}$.

$A$ 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 $\triangle OAB$ is

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