If the line bx + ay = 3ab intersects the coor | vedclass

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Question

(A) The given equation of the line is $bx + ay = 3ab$. Dividing both sides by $3ab$,we get: $\frac{bx}{3ab} + \frac{ay}{3ab} = 1$ $\frac{x}{3a} + \fra

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$(a, b)$

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(A) The given equation of the line is $bx + ay = 3ab$. Dividing both sides by $3ab$,we get: $\frac{bx}{3ab} + \frac{ay}{3ab} = 1$ $\frac{x}{3a} + \frac{y}{3b} = 1$. This is the intercept form of the line,where the $x$-intercept is $3a$ and the $y$-intercept is $3b$. Thus,the coordinates of the vertices of $\Delta OAB$ are $O(0, 0)$,$A(3a, 0)$,and $B(0, 3b)$. The centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$. Substituting the coordinates: $G = (\frac{0+3a+0}{3}, \frac{0+0+3b}{3}) = (\frac{3a}{3}, \frac{3b}{3}) = (a, b)$.

If the line $bx + ay = 3ab$ intersects the coordinate axes at points $A$ and $B$,then the centroid of $\Delta OAB$ is:

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