JEE MAIN - Mathematics (2024 - 31st January Morning Shift - No. 4)
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 $$\mathrm{ABCD}$$. If $$A B=\sqrt{10}$$ 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
8
5
12
10
Explanation
_31st_January_Morning_Shift_en_4_1.png)
Let E is mid point of diagonals
$$\frac{\alpha+\gamma}{2}=\frac{1+1}{2}$$
$$\alpha+\gamma=2$$
& $$\frac{\beta+\delta}{2}=\frac{2+0}{2}$$
$$\beta+\delta=2$$
$$2(\alpha+\beta+\gamma+\delta)=2(2+2)=8$$
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