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(A) Let the equation of the line passing through $(a, b)$ be $\frac{x}{h} + \frac{y}{k} = 1$. Since it passes through $(a, b)$,we have $\frac{a}{h} +

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$bx + ay - 3xy = 0$

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(A) Let the equation of the line passing through $(a, b)$ be $\frac{x}{h} + \frac{y}{k} = 1$. Since it passes through $(a, b)$,we have $\frac{a}{h} + \frac{b}{k} = 1$. The coordinates of $A$ are $(h, 0)$ and $B$ are $(0, k)$. The centroid $(x, y)$ of $	riangle OAB$ is given by $x = \frac{h+0+0}{3} = \frac{h}{3}$ and $y = \frac{0+k+0}{3} = \frac{k}{3}$. Thus,$h = 3x$ and $k = 3y$. Substituting these into the equation $\frac{a}{h} + \frac{b}{k} = 1$,we get $\frac{a}{3x} + \frac{b}{3y} = 1$. Multiplying by $3xy$,we obtain $ay + bx = 3xy$,or $bx + ay - 3xy = 0$.

$A$ variable straight line passes through a fixed point $(a, b)$ intersecting the coordinate axes at $A$ and $B$. If $O$ is the origin,then the locus of the centroid of the triangle $OAB$ is:

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