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JEE Mathematics 2025 Question with Solution

Let A(4,−2),B(1,1)A(4, -2), B(1, 1) and C(9,−3)C(9, -3) be the vertices of a triangle ABCABC. Then the maximum area of the parallelogram AFDEAFDE, formed with vertices D,ED, E and FF on the sides BC,CABC, CA and ABAB of the triangle ABCABC respectively, is _____.

  • A

    44

  • B

    66

  • C

    33

  • D

    99

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Triangle ABCABC has vertices A(4,−2),B(1,1),C(9,−3)A(4,-2), B(1,1), C(9,-3). A parallelogram AFDEAFDE is formed with F∈AB,E∈CA,D∈BCF \in AB, E \in CA, D \in BC.

Find: The maximum possible area of parallelogram AFDEAFDE.

First compute the area of triangle ABCABC using the determinant formula:

Area of △ABC=12∣4−211119−31∣\text{Area of } \triangle ABC = \frac{1}{2} \left| \begin{array}{ccc} 4 & -2 & 1 \\ 1 & 1 & 1 \\ 9 & -3 & 1 \end{array} \right| =6= 6

So, the area of triangle ABCABC is 66 square units.

For a parallelogram formed in this way with one vertex at AA and the other three vertices on the sides of triangle ABCABC, the maximum area is half the area of the triangle.

Therefore,

Maximum area of AFDE=12×6=3\text{Maximum area of } AFDE = \frac{1}{2} \times 6 = 3

Hence, the maximum area is 33 square units, so the correct option is C.

Vector Parametrization

Given: A(4,−2),B(1,1),C(9,−3)A(4,-2), B(1,1), C(9,-3).

Find: The maximum area of parallelogram AFDEAFDE with F∈AB,E∈AC,D∈BCF \in AB, E \in AC, D \in BC.

Let

u⃗=AB→=B−A=(1−4,  1−(−2))=(−3, 3)\vec{u}=\overrightarrow{AB}=B-A=(1-4,\;1-(-2))=(-3,\,3)

and

v⃗=AC→=C−A=(9−4,  −3−(−2))=(5, −1).\vec{v}=\overrightarrow{AC}=C-A=(9-4,\;-3-(-2))=(5,\,-1).

Now compute the determinant magnitude:

∣u⃗×v⃗∣=∣det⁡(−353−1)∣\left| \vec{u}\times\vec{v} \right|=\left| \det\begin{pmatrix} -3 & 5\\ 3 & -1\end{pmatrix} \right| =∣(−3)(−1)−3⋅5∣=∣3−15∣=12.=\left|(-3)(-1)-3\cdot 5\right|=|3-15|=12.

Hence,

[△ABC]=12∣u⃗×v⃗∣=12⋅12=6.[\triangle ABC]=\frac{1}{2}\left|\vec{u}\times\vec{v}\right|=\frac{1}{2}\cdot 12=6.

Take

F=A+αu⃗,E=A+βv⃗F=A+\alpha\vec{u}, \qquad E=A+\beta\vec{v}

where α,β∈[0,1]\alpha,\beta \in [0,1]. Then the fourth vertex of the parallelogram is

D=F+E−A=A+αu⃗+βv⃗.D=F+E-A=A+\alpha\vec{u}+\beta\vec{v}.

For DD to lie on side BCBC, we must have

α+β=1,α,β≥0.\alpha+\beta=1, \qquad \alpha,\beta \ge 0.

The area of the parallelogram is

[AFDE]=∣AF→×AE→∣=∣(αu⃗)×(βv⃗)∣=αβ∣u⃗×v⃗∣.[AFDE]=\left|\overrightarrow{AF}\times\overrightarrow{AE}\right|=\left|(\alpha\vec{u})\times(\beta\vec{v})\right|=\alpha\beta\left|\vec{u}\times\vec{v}\right|.

So,

[AFDE]=12αβ.[AFDE]=12\alpha\beta.

Now maximize αβ\alpha\beta subject to α+β=1\alpha+\beta=1. By AM-GM, the maximum occurs at

α=β=12\alpha=\beta=\frac{1}{2}

which gives

αβ=14.\alpha\beta=\frac{1}{4}.

Therefore,

[AFDE]max⁡=12⋅14=3.[AFDE]_{\max}=12\cdot\frac{1}{4}=3.

Thus the maximum area of the parallelogram is 33 square units, so the correct option is C.

Common mistakes

  • Using the area of the triangle directly as the area of the parallelogram is incorrect. The required figure is an inscribed parallelogram, and its maximum area is only half the area of the triangle. First find [△ABC][\triangle ABC], then take its half.

  • Making an error in the determinant calculation for [△ABC][\triangle ABC] leads to the wrong final answer. While computing the determinant or cross product, keep the coordinate order consistent and take the absolute value at the end.

  • Assuming any arbitrary points on ABAB and ACAC will work is incorrect. For the fourth vertex DD to lie on BCBC, the parameters must satisfy α+β=1\alpha+\beta=1. This condition is essential before maximizing the area.

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