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MCQEasyJEE 2026Straight Line Equations

JEE Mathematics 2026 Question with Solution

Rhombus vertices A(1,21,2), C(−3,−6-3,-6). Line AD parallel to 7x−y=147x-y=14. Find ∣α+β+γ+δ∣|\alpha+\beta+\gamma+\delta|.

  • A

    66

  • B

    11

  • C

    99

  • D

    33

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Rhombus vertices are A (1,2)\,(1,2) and C (−3,−6)\,(-3,-6). The required value is ∣α+β+γ+δ∣|\alpha+\beta+\gamma+\delta|.

Find: The magnitude of the sum of the coordinates of the remaining two vertices.

In a rhombus, the diagonals bisect each other, so the midpoint of AC is also the midpoint of BD.

M=(1+(−3)2,2+(−6)2)=(−1,−2)M = \left(\frac{1+(-3)}{2},\frac{2+(-6)}{2}\right) = (-1,-2)

Hence, for the other two vertices B(α,β)B(\alpha,\beta) and D(γ,δ)D(\gamma,\delta), we use the parallelogram property:

xA+xC=xB+xDx_A+x_C = x_B+x_D

and

yA+yC=yB+yDy_A+y_C = y_B+y_D

So,

α+γ=1+(−3)=−2\alpha+\gamma = 1+(-3) = -2

and

β+δ=2+(−6)=−4\beta+\delta = 2+(-6) = -4

Therefore,

α+β+γ+δ=(α+γ)+(β+δ)=−2+(−4)=−6\alpha+\beta+\gamma+\delta = (\alpha+\gamma) + (\beta+\delta) = -2 + (-4) = -6

Hence,

∣α+β+γ+δ∣=6|\alpha+\beta+\gamma+\delta| = 6

Therefore, the correct option is A.

Coordinate Sum Shortcut

Given: Opposite vertices of a rhombus are A (1,2)\,(1,2) and C (−3,−6)\,(-3,-6).

Find: ∣α+β+γ+δ∣|\alpha+\beta+\gamma+\delta|.

A rhombus is a parallelogram, so the sum of coordinates of one pair of opposite vertices equals the sum of coordinates of the other pair of opposite vertices.

Thus,

(xA+xC)+(yA+yC)=(α+γ)+(β+δ)(x_A+x_C) + (y_A+y_C) = (\alpha+\gamma) + (\beta+\delta)

Compute directly:

(1−3)+(2−6)=−2−4=−6(1-3) + (2-6) = -2-4 = -6

So,

α+β+γ+δ=−6\alpha+\beta+\gamma+\delta = -6

Hence,

∣α+β+γ+δ∣=6|\alpha+\beta+\gamma+\delta| = 6

Therefore, the correct option is A.

Common mistakes

  • Assuming the line AD parallel to 7x−y=147x-y=14 must be used. It is not needed here because the midpoint property of a parallelogram already gives the required sum directly. Use diagonal bisection first.

  • Treating AA and CC as adjacent vertices. They are opposite vertices here, so their midpoint is the intersection point of the diagonals. Using them as adjacent vertices leads to wrong coordinate relations.

  • Forgetting the absolute value at the end. The sum comes out to −6-6, but the question asks for ∣α+β+γ+δ∣|\alpha+\beta+\gamma+\delta|, so the required answer is 66.

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