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

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Straight Lines

Let

α, β, γ, δ \in Z

and let

A (α, β), B (1, 0), C (γ, δ)

and

D (1, 2)

be the vertices of a parallelogram ABCD . If

AB = 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

(A)10
(B)5
(C)12
(D)8✓ Correct

Answer: (D)

Solution

E

, the common midpoint of the diagonals

A C

and

B D

of parallelogram

A B C D

, satisfies

\frac{α + γ}{2} = \frac{1 + 1}{2} = 1 ⟹ α + γ = 2, \frac{β + δ}{2} = \frac{0 + 2}{2} = 1 ⟹ β + δ = 2

α + β + γ + δ = 2 + 2 = 4

, and

2 (α + β + γ + δ) = 8

(This is consistent with the additional constraints

A B = 10

and

A, C

3 y = 2 x + 1

, which fix the individual integer values of

α, β, γ, δ

without changing this sum.)

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option

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