Find the area of the parallelogram with vertices at (−5,3),(−5,3), (−1,1),(−1,1), (3,−6),(3,−6), and (7,−8)

Answer

46.5205

Q: Find the distance between points (-5,3) and (-1,1). A: Distance = sqrt(((-1)-(-5))^2 + ((1)-(3))^2) = 4.4721.
Q: Find the distance between points (-5,3) and (3,-6). A: Distance = sqrt(((3)-(-5))^2 + ((-6)-(3))^2) = 11.4018.
Q: Calculate the area of the parallelogram. A: Area = Distance1 * Distance2 * sin(angle), where the angle is the angle between the two sides formed by the vertices (-5,3), (-1,1), and (3,-6). To find the angle, we can use the dot product formula: cos(theta) = (a dot b) / (|a| * |b|), where a = (-1--5, 1-3) = (4,-2) and b = (3--1, -6-1) = (4,-7). Therefore, cos(theta) = (4*-4 + -2*-7) / (sqrt(4^2 + -2^2) * sqrt(4^2 + -7^2)) = (-30) / (sqrt(20) * sqrt(65)) = -0.3846. Since the parallelogram is not degenerate, we know that 0 < theta < pi, so we can take the absolute value of cos(theta) to get sin(theta) = sqrt(1 - cos(theta)^2) = sqrt(1 - 0.3846^2) = 0.9231. Therefore, Area = 4.4721 * 11.4018 * 0.9231 = 46.5205.