Given a parallelogram $ABCD$ lies in the $xy$ plane. The coordinates of point $B$ are $(-3,-4)$ and the coordinates of point $C$ are $(-7,-7)$, and $A$ are $(0,0)$. What is the area of the parallelogram?
This is a GRE practice question and I've struggled to answer it.
I am also a bit confused about this solution: Area of parallelogram = 2 * Triangle ABD. They say that if one of the vertices of a triangle is at the origin and the other two being $(a,b)$, $(c,d)$, then the area of the triangle can be written as:
Area of triangle $ABD = |\frac{ad-bc}{2}| = |\frac{(-3)(-7) - (-4)(-7)}{2}| = \frac{7}{2}.$
So the area of parallelogram is $7$.
Can anyone please explain to me how they got this formula? Also, are there alternate methods to solve this problem?
Given two vectors $u=(a,b), v=(c,d)$ in $R^2$ the area of the parallelogram with sides u and v is given by the abcolute value of the determinant
$$\begin{vmatrix} a&b\\c&d \end{vmatrix}=|ad-bc|$$
and the area of the triangle with sides u and v is given by
$$\frac12\begin{vmatrix} a&b\\c&d \end{vmatrix}=\frac12|ad-bc|$$