Let $\mathrm{A}(1,2)$ and $\mathrm{C}(-3,-6)$ be two diagonally opposite vertices of a rhombus, whose sides AD and BC are parallel to the line $7 \mathrm{x}-\mathrm{y}=14$. If $\mathrm{B}(\alpha, \beta)$ and $\mathrm{D}(\gamma, \delta)$ are the other two vertices, then $|\alpha+\beta+\gamma+\delta|$ is equal to :

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Straight Lines

Let

A (1, 2)

and

C (- 3, - 6)

be two diagonally opposite vertices of a rhombus, whose sides AD and BC are parallel to the line

7 x - y = 14

. If

B (α, β)

and

D (γ, δ)

are the other two vertices, then

∣ α + β + γ + δ ∣

is equal to :

(A)9
(B)3
(C)6✓ Correct
(D)1

Answer: (C)

Solution

In a rhombus the diagonals bisect each other, so

B D

and

A C

share the same midpoint. The midpoint of

A C

(\frac{1 - 3}{2}, \frac{2 - 6}{2}) = (- 1, - 2)

. Hence

\frac{α + γ}{2} = - 1

and

\frac{β + δ}{2} = - 2

, giving

α + γ = - 2

and

β + δ = - 4

. Therefore

∣ α + β + γ + δ ∣ = ∣ - 2 - 4∣ = 6

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option

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