Let A(-2,-1),B(1,0),C(alpha, beta),and D(gamm | vedclass

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(C) In a parallelogram,the diagonals bisect each other. Let $P$ be the midpoint of $AC$ and $BD$.$P = \left(\frac{\alpha-2}{2}, \frac{\beta-1}{2}\righ

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$32$

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(C) In a parallelogram,the diagonals bisect each other. Let $P$ be the midpoint of $AC$ and $BD$.$P = \left(\frac{\alpha-2}{2}, \frac{\beta-1}{2}\right) = \left(\frac{\gamma+1}{2}, \frac{\delta+0}{2}\right)$Equating the coordinates,we get:$\alpha-2 = \gamma+1 \Rightarrow \alpha-\gamma = 3 \dots(1)$$\beta-1 = \delta \Rightarrow \beta-\delta = 1 \dots(2)$Since $C(\alpha, \beta)$ lies on $2x-y=5$,we have $2\alpha-\beta=5 \dots(3)$Since $D(\gamma, \delta)$ lies on $3x-2y=6$,we have $3\gamma-2\delta=6 \dots(4)$From $(2)$,$\delta = \beta-1$. Substituting into $(4)$:$3\gamma - 2(\beta-1) = 6 \Rightarrow 3\gamma - 2\beta = 4 \dots(5)$From $(1)$,$\gamma = \alpha-3$. Substituting into $(5)$:$3(\alpha-3) - 2\beta = 4$ $\Rightarrow 3\alpha - 9 - 2\beta = 4$ $\Rightarrow 3\alpha - 2\beta = 13 \dots(6)$Solving $(3)$ and $(6)$:$2\alpha - \beta = 5 \Rightarrow \beta = 2\alpha - 5$$3\alpha - 2(2\alpha-5) = 13$ $\Rightarrow 3\alpha - 4\alpha + 10 = 13$ $\Rightarrow -\alpha = 3$ $\Rightarrow \alpha = -3$$\beta = 2(-3) - 5 = -11$$\gamma = -3-3 = -6$$\delta = -11-1 = -12$$|\alpha+\beta+\gamma+\delta| = |-3-11-6-12| = |-32| = 32$

Let $A(-2,-1)$,$B(1,0)$,$C(\alpha, \beta)$,and $D(\gamma, \delta)$ be the vertices of a parallelogram $ABCD$. If the point $C$ lies on $2x-y=5$ and the point $D$ lies on $3x-2y=6$,then the value of $|\alpha+\beta+\gamma+\delta|$ is equal to . . . . . . .

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