1. Set up the triangle and its area

Given: $A(4,-2),\quad B(1,1),\quad C(9,-3)$

First, find the area of $\triangle ABC$.

Using vectors from $A$: $\overrightarrow{AB}=(1-4,1-(-2))=(-3,3)$ $\overrightarrow{AC}=(9-4,-3-(-2))=(5,-1)$

Area of triangle: $[ABC]=\frac12\left|\begin{vmatrix}-3 & 3\\ 5 & -1\end{vmatrix}\right|$

Compute determinant: $(-3)(-1)-3(5)=3-15=-12$ So, $[ABC]=\frac12|{-12}|=6$

2. Use a general property of an inscribed parallelogram

Parallelogram $AFDE$ has:

• $A$ as one vertex,
• $F$ on $AB$,
• $E$ on $AC$,
• $D$ on $BC$.

Let $\overrightarrow{AF}=t\,\overrightarrow{AB}, \qquad \overrightarrow{AE}=s\,\overrightarrow{AC}$ for some $0\le t,s\le 1$.

Since $AFDE$ is a parallelogram with adjacent sides $AF$ and $AE$, the fourth vertex is $D=A+\overrightarrow{AF}+\overrightarrow{AE}$

For $D$ to lie on side $BC$, it must be representable as $D=(1-\lambda)B+\lambda C$

Now write $D$ in barycentric form using $A,B,C$: $D=A+t(B-A)+s(C-A)=(1-t-s)A+tB+sC$

A point on line segment $BC$ has coefficient of $A$ equal to $0$. Hence, $1-t-s=0 \quad \Rightarrow \quad t+s=1$

3. Area of the parallelogram

Area of parallelogram $AFDE$ is $[AFDE]=|\overrightarrow{AF}\times \overrightarrow{AE}|$

Since $\overrightarrow{AF}=t\overrightarrow{AB}, \qquad \overrightarrow{AE}=s\overrightarrow{AC},$ we get $[AFDE]=ts\,|\overrightarrow{AB}\times \overrightarrow{AC}|$

But $|\overrightarrow{AB}\times \overrightarrow{AC}|=2[ABC]=12$ So, $[AFDE]=12ts$

Using the condition $t+s=1$, let $s=1-t$. Then $[AFDE]=12t(1-t)$

4. Maximize the area

Consider $f(t)=12t(1-t)=12(t-t^2), \qquad 0\le t\le 1$

Differentiate: $f'(t)=12(1-2t)$ Set $f'(t)=0$: $1-2t=0 \Rightarrow t=\frac12$ Then $s=1-t=\frac12$

Maximum area: $[AFDE]_{\max}=12\cdot \frac12\cdot \frac12=3$

5. Final answer

The maximum area of the parallelogram is $\boxed{3}$

6. Comparison with stored answer

Stored correct answer: $3$

Our derived answer is also $3$, so it agrees.