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(D) In a parallelogram,the diagonals bisect each other. Therefore,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.The midpo

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$8$

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(D) In a parallelogram,the diagonals bisect each other. Therefore,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.The midpoint of $BD$ is $\left(\frac{1+1}{2}, \frac{2+0}{2}\right) = (1, 1)$.The midpoint of $AC$ is $\left(\frac{\alpha+\gamma}{2}, \frac{\beta+\delta}{2}\right)$.Equating the midpoints,we get:$\frac{\alpha+\gamma}{2} = 1 \implies \alpha + \gamma = 2$$\frac{\beta+\delta}{2} = 1 \implies \beta + \delta = 2$We need to find the value of $2(\alpha + \beta + \gamma + \delta)$.Substituting the sums:$2(\alpha + \gamma + \beta + \delta) = 2(2 + 2) = 2(4) = 8$.

Let $\alpha, \beta, \gamma, \delta \in \mathbb{Z}$ and let $A(\alpha, \beta), B(1, 0), C(\gamma, \delta)$ and $D(1, 2)$ be the vertices of a parallelogram $ABCD$. If $AB = \sqrt{10}$ and the points $A$ and $C$ lie on the line $3y = 2x + 1$,then $2(\alpha + \beta + \gamma + \delta)$ is equal to

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