Let alpha, beta, gamma, delta in Z$\alpha, \beta, \gamma, \delta \in Z$ and let A (alpha, beta)$A (\alpha, \beta)$, B (1, 0)$B (1, 0)$, C (gamma, delta)$C (\gamma, \delta)$ and D (1, 2)$D (1, 2)$ be the vertices of a parallelogram ABCD$ABCD$. If AB = sqrt10$AB = \sqrt{10}$ and the points A$A$ and C$C$ lie on the line 3y = 2x + 1$3y = 2x + 1$, then 2(alpha + beta + gamma + delta)$2(\alpha + \beta + \gamma + \delta)$ is equal to

### Core Logic Let E$E$ be the midpoint of the diagonals AC$AC$ and BD$BD$. Since ABCD$ABCD$ is a parallelogram, the diagonals bisect each other.

Midpoint E$E$ from BD$BD$: left( frac1+12, frac0+22 right) = (1, 1)$\left( \frac{1+1}{2}, \frac{0+2}{2} \right) = (1, 1)$. ### Step 1: Apply Midpoint on AC Midpoint E$E$ from AC$AC$: left( fracalpha+gamma2, fracbeta+delta2 right)$\left( \frac{\alpha+\gamma}{2}, \frac{\beta+\delta}{2} \right)$. Equating both: fracalpha+gamma2 = 1 implies alpha + gamma = 2$$\frac{\alpha+\gamma}{2} = 1 \implies \alpha + \gamma = 2$$ fracbeta+delta2 = 1 implies beta + delta = 2$$\frac{\beta+\delta}{2} = 1 \implies \beta + \delta = 2$$ ### Step 2: Final Value The expression requires 2(alpha + beta + gamma + delta)$2(\alpha + \beta + \gamma + \delta)$. 2(2 + 2) = 2(4) = 8$$2(2 + 2) = 2(4) = 8$$ *(Note: Additional conditions like AB = sqrt10$AB = \sqrt{10}$ and the line equation are extraneous data not needed to find the sum).* ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines