Find the area of the parallelogram with vertices (3,2), (5, 6), (9, 11), and (11, 15)

area-parallelogram

#1

Find the area of the parallelogram with vertices (3,2), (5, 6), (9, 11), and (11, 15).

Answer:

As we have the formula, the area of a parallelogram created by the extension of two vectors with the same endpoint is equal to the absolute value of the determinant of vector 1 on top of vector 2 i.e. area=|x1y2-x2y1|
Lets set vector 1 = (x1,y1) and vector 2 = (x2,y2)
here we can pick any point on the parallelogram and the two vectors extending from that point will be the sides of the parallelogram.

Take (3,2) as the starting point. The endpoints of vector 1 and vector 2 will be (5,6) and (9,11) respectively as those vectors make the two sides of the parallelogram.

Vector 1 = (5-3,6-2) = (2,4)
Vector 2 = (9-3,11-2) = (6,9)
Area = |29 - 46|= |18-24| = 6