Let A(4, -2),B(1, 1),and C(9, -3) be the vert | vedclass

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(D) The area of $	riangle ABC$ is given by the determinant formula: $	ext{Area} = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$.$\

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$3$

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(D) The area of $	riangle ABC$ is given by the determinant formula: $	ext{Area} = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$.$	ext{Area} = \frac{1}{2} |4(1 - (-3)) + 1(-3 - (-2)) + 9(-2 - 1)|$.$	ext{Area} = \frac{1}{2} |4(4) + 1(-1) + 9(-3)| = \frac{1}{2} |16 - 1 - 27| = \frac{1}{2} |-12| = 6$ square units.For a parallelogram $AFDE$ inscribed in a triangle $ABC$ such that $D$ lies on $BC$,$E$ on $AC$,and $F$ on $AB$,the maximum area of the parallelogram is exactly half the area of the triangle $ABC$.$	ext{Maximum Area} = \frac{1}{2} 	imes 	ext{Area}(	riangle ABC) = \frac{1}{2} 	imes 6 = 3$ square units.

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 $\qquad$ .

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