# Find the area of the parallelogram with vertices (1,2), (4, 5), (5, 9), and (8, 12)

1. Sep 21, 2008

### samazing18

i'm posting this problem in the calculus forum because i got this question in a calculus class. It seems like a straightforward area problem, but i don't think that's the case and i can't figure out another way to do the problem using vectors.

"Find the area of the parallelogram with vertices (1,2), (4, 5), (5, 9), and (8, 12)."

any ideas?

2. Sep 21, 2008

### JG89

Re: Find the area of the parallelogram with vertices (1,2), (4, 5), (5, 9), and (8, 1

Add in a third coordinate of 0 and then find the magnitude of the cross product of two adjacent sides, since the cross product isn't defined in R2.

Last edited: Sep 21, 2008
3. Sep 21, 2008

### JG89

Re: Find the area of the parallelogram with vertices (1,2), (4, 5), (5, 9), and (8, 1

If you know the answer already, try that out and see if it works.

4. Sep 21, 2008

### JG89

Re: Find the area of the parallelogram with vertices (1,2), (4, 5), (5, 9), and (8, 1

Another way I thought of that doesn't involve the cross product would be to find the magnitude of the vector forming the base. Then find the magnitude of a vector that would be the height of the parallelogram. To do this, take one of the top vertices, and find a vector that is orthogonal to it and find the point at which that vector (the orthogonal one) intersects the other vector. Take the magnitude of the vector from that point of intersection to the original vertice and you should have the height. Then the area of the parallelogram is just base times height.

5. Sep 22, 2008

### uocan

Re: Find the area of the parallelogram with vertices (1,2), (4, 5), (5, 9), and (8, 1

another, and perhaps simpler, ways is to take the difference of the inner product of adjacent vectors and multiplication of magnitudes of the adjacent vectors

Last edited: Sep 22, 2008