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(C) The vertex $A$ is the intersection of $2x - 3y = -23$ and $3x + 7y = 23$. Solving these,we get $A = (-4, 5)$.The vertex $C$ is the intersection of

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$529$

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(C) The vertex $A$ is the intersection of $2x - 3y = -23$ and $3x + 7y = 23$. Solving these,we get $A = (-4, 5)$.The vertex $C$ is the intersection of $5x + 4y = 23$ and $3x + 7y = 23$. Solving these,we get $C = (3, 2)$.The midpoint of diagonal $AC$ is $M = \left(\frac{-4+3}{2}, \frac{5+2}{2}ight) = \left(-\frac{1}{2}, \frac{7}{2}ight)$.Since the diagonals of a parallelogram bisect each other,the other diagonal $BD$ passes through $M$ and the intersection of the other two sides $B$ (intersection of $2x - 3y = -23$ and $5x + 4y = 23$),which is $B = (-1, 7)$.The slope of $BD$ is $m = \frac{7 - 7/2}{-1 - (-1/2)} = \frac{7/2}{-1/2} = -7$.The equation of diagonal $BD$ is $y - 7 = -7(x + 1)$,which simplifies to $7x + y = 0$.The distance $d$ of point $A(-4, 5)$ from the line $7x + y = 0$ is $d = \frac{|7(-4) + 5|}{\sqrt{7^2 + 1^2}} = \frac{|-28 + 5|}{\sqrt{50}} = \frac{23}{\sqrt{50}}$.Therefore,$50d^2 = 50 	imes \left(\frac{23}{\sqrt{50}}ight)^2 = 50 	imes \frac{529}{50} = 529$.

Let the equations of two adjacent sides of a parallelogram $ABCD$ be $2x - 3y = -23$ and $5x + 4y = 23$. If the equation of its one diagonal $AC$ is $3x + 7y = 23$ and the distance of $A$ from the other diagonal is $d$,then $50d^2$ is equal to $........$.

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