If the line 2x - 3y + 4 = 0 divides the line | vedclass

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(A) Let the point $P$ divide the segment $AB$ in the ratio $m:n$. By the section formula,the coordinates of $P$ are $\left(\frac{3m - 2n}{m + n}, \fra

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$(-17, 18)$

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(A) Let the point $P$ divide the segment $AB$ in the ratio $m:n$. By the section formula,the coordinates of $P$ are $\left(\frac{3m - 2n}{m + n}, \frac{-2m + 3n}{m + n}ight)$.Since $P$ lies on the line $2x - 3y + 4 = 0$,we have $2\left(\frac{3m - 2n}{m + n}ight) - 3\left(\frac{-2m + 3n}{m + n}ight) + 4 = 0$.Multiplying by $(m + n)$,we get $6m - 4n + 6m - 9n + 4m + 4n = 0$,which simplifies to $16m - 9n = 0$,so $\frac{m}{n} = \frac{9}{16}$.We need to find the point dividing $AB$ in the ratio $k = \frac{-4m}{3n} = \frac{-4}{3} 	imes \frac{9}{16} = -\frac{3}{4}$.Using the section formula for ratio $k = -\frac{3}{4}$ with $A(-2, 3)$ and $B(3, -2)$:$x = \frac{k x_2 + x_1}{k + 1} = \frac{-\frac{3}{4}(3) + (-2)}{-\frac{3}{4} + 1} = \frac{-\frac{9}{4} - 2}{\frac{1}{4}} = -9 - 8 = -17$.$y = \frac{k y_2 + y_1}{k + 1} = \frac{-\frac{3}{4}(-2) + 3}{-\frac{3}{4} + 1} = \frac{\frac{3}{2} + 3}{\frac{1}{4}} = 4 	imes \frac{9}{2} = 18$.Thus,the point is $(-17, 18)$.

If the line $2x - 3y + 4 = 0$ divides the line segment joining the points $A(-2, 3)$ and $B(3, -2)$ in the ratio $m:n$,then the point which divides $AB$ in the ratio $-4m:3n$ is

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