JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Straight Lines

Let

A (4, - 2), B (1, 1)

and

C (9, - 3)

be the vertices of a triangle ABC. Then the maximum area of the parallelogram AFDE, formed with vertices D, E and F on the sides BC, CA and AB of the triangle ABC respectively, is ______.

Answer: 3

Solution

The area of triangle

A B C

\frac{1}{2} 4 (1 + 3) + 1 (- 3 + 2) + 9 (- 2 - 1) = \frac{1}{2} ∣16 - 1 - 27∣ = 6

. For a parallelogram

A F D E

with vertex

A

and the other vertices on the sides, taking

F

A B

and

E

A C

dividing those sides in the same ratio

t : (1 - t)

makes the parallelogram's area equal to

2 t (1 - t)

times the triangle's area; this is greatest at

t = \frac{1}{2}

, where it equals half the triangle. Hence the maximum area is

\frac{1}{2} \cdot 6 = 3

square units.

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Integer answer

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