Find the area of the parallelogram with vertices A(−1,3,3), B(0,5,7), C(1,2,6), and D(2,4,10).

Step-by-step explanation:

The diagonal of the parallelogram ABCD divides it into 2 equal triangles. Considering triangle ABC, it means that the area of the parallelogram would be

2 × area of triangle ABC

Writing the vertices of triangle ABC,

A(−1,3,3), B(0,5,7), C(1,2,6)

We would determine the length of each side of the triangle.

AB = √(0 - - 1)² + (5 - 3)² + (7 - 3)^2

AB = √(1 + 4 + 16) = √21

BC = √(1 - 0)² + (2 - 5)² + (6 - 7)²

BC = √(1 + 9 + 1) = √11

AC = √(1 - - 1)² + (2 - 3)² + (6 - 3)²)

AC = √(4 + 1 + 9) = √14

We would apply the heron's formula for determining the area of a triangle

Area = √s(s - a)(s - b)(s - c)

Where

s = (a + b + c)/2

a = AB, b = BC, c = AC

s = (√21 + √11 + √14)/2 = 5.82

s - a = 5.82 - √21 = 1.24

s - b = 5.82 - √11 = 2.5

s - c = 5.82 - √14 = 2.08

Area = √(5.82 × 1.24 × 2.5 × 2.08) = 6.126

Therefore, area of parallelogram ABCD is

6.126 × 2 = 12.252

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )