solution

Solution: part a) Since the area of the the triangle ABC is half of the area of the parallelogram ABCD, the area of the triangle ABC is halfe of the determinant of the two vectors formin the parallelogram. The two vectors forming the parallelogram are ${\bf u } = AB= ( -3-1, 4-2)= ( -4, 2)$ and ${\bf v} = AC = ( 2-1, 4-2)= ( 1, 2)$.

Hence Area of ABC = $\frac{ 1} {2}$ area of the parallelogram ABCD $= \frac{ 1} {2}\left\vert \det\left(\left[ \begin{array}{rr} -4 & 2\\ 1&2\\ \end {array} \right]\right) \right\vert=\frac{ 1} {2}\vert-10 \vert = 5$.

Solution: part b) The area of the parallelogram ABCD is 10 twice of the area of the triangle ABC.