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(C) Let the parallelogram be $ABCD$. The lines $4x + 5y = 0$ and $7x + 2y = 0$ intersect at the origin $A(0, 0)$.Let the diagonal $BD$ lie on the line

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$(2, 2)$

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(C) Let the parallelogram be $ABCD$. The lines $4x + 5y = 0$ and $7x + 2y = 0$ intersect at the origin $A(0, 0)$.Let the diagonal $BD$ lie on the line $11x + 7y = 9$.Point $D$ is the intersection of $4x + 5y = 0$ and $11x + 7y = 9$. Solving these,we get $D = (5/3, -4/3)$.Point $B$ is the intersection of $7x + 2y = 0$ and $11x + 7y = 9$. Solving these,we get $B = (-2/3, 7/3)$.The diagonals of a parallelogram bisect each other. Let $M$ be the midpoint of $BD$.$M = (\frac{5/3 - 2/3}{2}, \frac{-4/3 + 7/3}{2}) = (1/2, 1/2)$.The other diagonal $AC$ passes through $A(0, 0)$ and $M(1/2, 1/2)$.The equation of line $AC$ is $y - 0 = \frac{1/2 - 0}{1/2 - 0}(x - 0)$,which simplifies to $y = x$.Checking the options,the point $(2, 2)$ satisfies $y = x$.

Two sides of a parallelogram are along the lines $4x + 5y = 0$ and $7x + 2y = 0$. If the equation of one of the diagonals of the parallelogram is $11x + 7y = 9$,then the other diagonal passes through the point:

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