Let $\mathrm{A}(4,-2), \mathrm{B}(1,1)$ and $\mathrm{C}(9,-3)$ be the vertices of a triangle $A B C$. Then…
Let $\mathrm{A}(4,-2), \mathrm{B}(1,1)$ and $\mathrm{C}(9,-3)$ be the vertices of a triangle $A B C$. Then the maximum area of the parallelogram AFDE , formed with vertices $\mathrm{D}, \mathrm{E}$ and F on the sides $\mathrm{BC}, \mathrm{CA}$ and AB of the triangle ABC respectively, is ______ .
Solution
Area of $\triangle \mathrm{ABC}=\frac{1}{2}\left|\begin{array}{ccc}4 & -2 & 1 \\ 1 & 1 & 1 \\ 9 & -3 & 1\end{array}\right|$ $=6$ square units Maximum area of $\operatorname{AFDE}=\frac{1}{2} \times 6=3$ sq. units