JEE Main 2025 — 2 April, Evening Shift — Question 46

The maximum area of such a parallelogram $A F D E$, with one vertex fixed at $A$ and the other three points

lying on the sides of triangle $A B C$, is half the area of triangle $A B C$.

Using the determinant formula for area of triangle with vertices $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)$ :

Area $\triangle A B C$

$=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

Substitute the coordinates: $=\frac{1}{2}|4(1-(-3))+1((-3)-(-2))+9((-2)-1)|$

$=\frac{1}{2}|4(4)+1(-1)+9(-3)|$

$=\frac{1}{2}|16-1-27|=\frac{1}{2}|-12|=\frac{12}{2}=6$

Maximum area of parallelogram $A F D E$

$=\frac{1}{2} \times$ area of triangle $=\frac{1}{2} \times 6=3$