Find the area of the parallelogram with vertices $(4,1)$, $(6, 6)$, $(7, 7)$, and $(9, 12)$.

Question

I am trying to find the area of the parallelogram with vertices $(4,1)$, $(6, 6)$, $(7, 7)$, and $(9, 12)$.

So I believe the way to solve this problem is through the cross product and then taking the magnitude. However, I got the wrong answer and I am now rethinking my methods.

My original work:

Point $A$: $(4,1)$; Point $B$: $(6,6)$; Point $C$: $(7,7)$; Point $D$: $(9,12)$.

I took the cross product of $\vec{AB}$ and $\vec{AC}$ and took the magnitude of that and ended up getting $\sqrt{369}$.

I would appreciate the help. I am kind of stumped.

Answer 1

See this illustration on how to compute the area of a parallelogram in an easy manner.

Answer 2

area should be 3 square units.