Let A(4,-2),B(1,1) and C(9,-3) be the vertices of a triangle ABC .

Question

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\_\_\_\_\_\_\_\_\_\_\_

Answer: 3

Accepted Answer

The maximum area of such a parallelogram $AFDE$, with one vertex fixed at $A$ and the other three points lying on the sides of triangle $ABC$, is half the area of triangle $ABC$.

Using the determinant formula for area of triangle with vertices $A({x}_{1},{y}_{1}),B({x}_{2},{y}_{2}),C({x}_{3},{y}_{3})$ :

Area $	riangleABC$

$=\frac{1}{2}|{x}_{1}({y}_{2}-{y}_{2}^{3})+{x}_{2}({y}_{3}-{y}_{1})+{x}_{3}({y}_{1}-{y}_{2})|$

Substitute the coordinates:

$\begin{gathered} =\frac{1}{2}|4(1-(-3))+1((-3)-(-2))+9((-2)-1)| \ =\frac{1}{2}|4(4)+1(-1)+9(-3)| \ =\frac{1}{2}|16-1-27|=\frac{1}{2}|-12|=\frac{12}{2}=6 \end{gathered}$

Maximum area of parallelogram $AFDE$ $=\frac{1}{2}	imes$ area of triangle $=\frac{1}{2}	imes 6=3$

Solutions · 1