Let A(α, β), B(1, 0), C(γ, δ) and D(1, 2) Be the Vertices of a Parallelogram ABCD With α, β, γ, δ ∈ Z. If AB = √10 and the Points A and C Lie on the Line 3y = 2x + 1, Then 2(α + β + γ + δ) Is Equal To

JEE Main 2024 — 31 January, Shift 1. Previous Year Question.

Problem

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

Key insight. A parallelogram’s diagonals bisect each other — so the midpoint of $BD$ (fully known, since both $B$ and $D$ are given) must equal the midpoint of $AC$, immediately pinning down $\alpha+\gamma$ and $\beta+\delta$ without ever needing the individual values of $\alpha$, $\beta$, $\gamma$, $\delta$.

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Approach

Notice that the quantity asked for, $\alpha+\beta+\gamma+\delta$, doesn’t require knowing $A$ and $C$ individually — just their coordinate sums. Since $ABCD$ is a parallelogram, its diagonals $AC$ and $BD$ bisect each other, so their midpoints coincide. Computing the midpoint of $BD$ (both endpoints fully known) and setting it equal to the midpoint of $AC$ gives $\alpha+\gamma$ and $\beta+\delta$ directly — the extra given conditions about $AB=\sqrt{10}$ and the line $3y=2x+1$ turn out not to be needed for this particular question.

Solution

Step 1 — Recall the diagonal-bisection property

In parallelogram $ABCD$, the diagonals are $AC$ and $BD$, and they bisect each other — their midpoints are the same point.

Step 2 — Find the midpoint of BD

With $B=(1,0)$ and $D=(1,2)$:

$$\text{Midpoint of } BD = \left(\frac{1+1}2,\frac{0+2}2\right) = (1,1)$$

Step 3 — Set the midpoint of AC equal to this

With $A=(\alpha,\beta)$ and $C=(\gamma,\delta)$:

$$\text{Midpoint of } AC = \left(\frac{\alpha+\gamma}2,\frac{\beta+\delta}2\right)$$

Setting this equal to $(1,1)$:

$$\frac{\alpha+\gamma}2=1 \implies \alpha+\gamma=2, \qquad \frac{\beta+\delta}2=1 \implies \beta+\delta=2$$

Step 4 — Compute the requested quantity

$$2(\alpha+\beta+\gamma+\delta) = 2\left[(\alpha+\gamma)+(\beta+\delta)\right] = 2(2+2) = 2\times4 = 8$$

Answer

Common mistakes

• Trying to use the condition $AB=\sqrt{10}$ and the line $3y=2x+1$ to solve for $\alpha,\beta,\gamma,\delta$ individually. These conditions genuinely do pin down the four integer values uniquely (up to the constraint of being integers), but doing so is unnecessary extra work — the question only asks for a specific sum of the four values, which the diagonal-bisection property gives directly.
• Mixing up which pair of vertices forms each diagonal. In parallelogram $ABCD$ (vertices in order), the diagonals are $AC$ and $BD$ — not $AB$ and $CD$, which are sides. Using the wrong pairing gives a midpoint condition that doesn’t correspond to any real property of the parallelogram.
• Forgetting the final multiplication by 2. It’s easy to correctly find $\alpha+\beta+\gamma+\delta=4$ and report that as the answer, overlooking that the question specifically asks for $2(\alpha+\beta+\gamma+\delta)=8$.

Practise next

• Let $\alpha,\beta,\gamma,\delta\in\mathbb Z$ and let $A(\alpha,\beta)$, $B(1,0)$, $C(\gamma,\delta)$, $D(2,3)$ be the vertices of a parallelogram $ABCD$. Find $2(\alpha+\beta+\gamma+\delta)$.