Let α , β , γ , δ ∈ Z and let A α , β , B 1 , 0 , C γ , δ and D 1 , 2 be the vertices of a parallelogram A B C D . If A B = 10 and the points A and C lie on the line 3 y = 2 x + 1 , then 2 α + β + γ +

Question

Let α , β , γ , δ ∈ Z and let A α , β , B 1 , 0 , C γ , δ and D 1 , 2 be the vertices of a parallelogram A B C D . If A B = 10 and the points A and C lie on the line 3 y = 2 x + 1 , then 2 α + β + γ + δ is equal to

TestHub · Accepted Answer

Plotting the diagram of the given data we get,

We know that diagonals of a parallelogram bisect each other.

$\Rightarrow Mid-point of A C = Mid-point of B D$

$\Rightarrow (\frac{α + γ}{2}, \frac{β + δ}{2}) = (\frac{1 + 1}{2}, \frac{0 + 2}{2})$

$\Rightarrow (\frac{α + γ}{2}, \frac{β + δ}{2}) = (1, 1)$

$\Rightarrow α + γ = 2, β + δ = 2$

$\Rightarrow 2 (α + γ + β + δ) = 8$

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