The line $b x + a y = 3 a b$ cuts the coordinate axes at $A$ and $B$ , then centroid of $△ O A B$ is -

(b, a)

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(a, b)

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(\frac{a}{3}, \frac{b}{3})

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(3 a, 3 b)

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Solution

The correct option is B $(a, b)$
REF.Image
$b x + a y = 3 a b$
$\Rightarrow \frac{b x + a y}{3 a b} = 1 \Rightarrow \frac{x}{3 a} + \frac{y}{3 b} = 1$
centroid $\Rightarrow (\frac{0 + 0 + 3 a}{3}, \frac{0 + 3 b + 0}{3})$
$\Rightarrow (a, b)$

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(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}

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