Question

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

• A. 10
• B. 5
• C. 12
• D. 8

Answer 1

The Correct Option is D

Let \( E \) be the midpoint of the diagonals. By the midpoint formula:

\[ \frac{\alpha + \gamma}{2} = \frac{1 + 1}{2} = 1 \quad \implies \quad \alpha + \gamma = 2 \]

Similarly:

\[ \frac{\beta + \delta}{2} = \frac{2 + 0}{2} = 1 \quad \implies \quad \beta + \delta = 2 \]

Therefore:

\[ 2(\alpha + \beta + \gamma + \delta) = 2(2 + 2) = 8 \]