JEE Main 2025 (Online) 2nd April Evening ShiftQuestion 1
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 $\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$3$
Solution
Concept. With $AF=\lambda\,AB$ and $AE=\mu\,AC$, the fourth vertex $D$ is on $BC$ exactly when $\lambda+\mu=1$; the area is $2\lambda\mu$ times the triangle. Formulas used. (1) Area of triangle with $\vec{AB}=(p,q)$, $\vec{AC}=(r,s)$: $\frac12|ps-qr|$(2) $\lambda+\mu=1$, $\lambda,\mu>0$ $\Rightarrow$ $\lambda\mu\le\frac14$Step 1 - Triangle. $\vec{AB}=(-3,3)$, $\vec{AC}=(5,-1)$. By (1), area $=\frac12|3-15|=6$. Step 2 - Parallelogram. $D=A+\lambda\vec{AB}+\mu\vec{AC}$ lies on $BC$ iff $\lambda+\mu=1$. Its area is $$|\lambda\vec{AB}\times\mu\vec{AC}|=\lambda\mu\cdot12=2\lambda\mu\cdot6.$$ Step 3 - Maximise. By (2), the largest value is at $\lambda=\mu=\frac12$ ($F,E,D$ are midpoints):$$\text{area}=2\cdot\frac14\cdot6=3.$$ Answer: 3