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9.Straight Line
hard
Let $\alpha, \beta, \gamma, \delta \in \mathrm{Z}$ and let $\mathrm{A}(\alpha, \beta), \mathrm{B}(1,0), \mathrm{C}(\gamma, \delta)$ and $D(1,2)$ be the vertices of a parallelogram $\mathrm{ABCD}$. If $\mathrm{AB}=\sqrt{10}$ and the points $\mathrm{A}$ and $\mathrm{C}$ lie on the line $3 y=2 x+1$, then $2(\alpha+\beta+\gamma+\delta)$ is equal to
A
$10$
B
$5$
C
$12$
D
$8$
(JEE MAIN-2024)
Solution

Let $\mathrm{E}$ is mid point of diagonals
$\begin{array}{ll}\frac{\alpha+\gamma}{2}=\frac{1+1}{2} & \& \frac{\beta+\delta}{2}=\frac{2+0}{2} \\ \alpha+\gamma=2 & \beta+\delta=2 \\ 2(\alpha+\beta+\gamma+\delta)=2(2+2)=8\end{array}$
Standard 11
Mathematics