Find the area of the parallelogram with vertices A(-3, 0), B(-1, 3), C(5, 2), and D(3, -1)

area-parallelogram

#1

Find the area of the parallelogram with vertices A(-3, 0), B(-1, 3), C(5, 2), and D(3, -1)

Answer:

Area of parallelogram = ab sin (theta)=IaxbI (cross product)
here a and b are two adjacent sides of the parallelogram. Theta is the angle between these two vertices.
Here a=(AB)=(-1,3)-(-3,0) = 2 i +3j
b=(AD)=(3,-1)-(-3,0)= 6i+3j
Here i and j are unit vectors in X and Y direction respectively
axb =(2 i +3j) x (6i+3j) = 20 k (k unit vector in z )
area of parallelogram =IaxbI =20 sq. units