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

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 AB and AC and took the magnitude of that and ended up getting sqrt(369).

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

• Probably AB or AC is a diagonal rather than a side. Draw a picture and see, otherwise your approach is good and the error is just an arithmetic mistake. – user142299 May 14 '14 at 0:40
• I drew it multiple times on a large paper and it was still very ambiguous. I am not sure if I am taking the right points...? – Rohit Tigga May 14 '14 at 0:42
• The cross product is much shorter than you have calculated. Did you subtract all the components of A from B to get the vector AB? Please show how you got $\sqrt {369}$ It is true that neither AB nor AC is a diagonal. – Ross Millikan May 14 '14 at 0:50

• Why?${}{}{}{}{}$ – vonbrand May 14 '14 at 1:21