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
A.10$10$
B.5$5$
C.12$12$
D.8$8$
Solution & Explanation
### 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.
Properties of Parallelogram diagram for Q10 - JEE Main 2024 Morning
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
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### Chapter Mix
Class 11 Maths: Straight Lines
Let A(-2, -1)$A(-2, -1)$, B(1, 0)$B(1, 0)$, C(alpha, beta)$C(\alpha, \beta)$ and D(gamma, delta)$D(\gamma, \delta)$ be the vertices of a parallelogram ABCD. If the point C lies on 2x - y = 5$2x - y = 5$ and the point D lies on 3x - 2y = 6$3x - 2y = 6$, then the value of |alpha + beta + gamma + delta|$|\alpha + \beta + \gamma + \delta|$ is equal to
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