Final answer:
The area of the parallelogram with vertices at A(-1, 5), B(3, 2), C(0, 2), and D(-13, -11) is 9 square units. This is obtained by using the coordinates to calculate vectors ΔAB and ΔAC and then taking the absolute value of the determinant of these two vectors.
Step-by-step explanation:
The area of a parallelogram can be determined by using the vertices provided to calculate the base and height, or directly using the coordinates to find the determinant of two vectors formed by the sides of the parallelogram. In this particular problem, we will find vectors ΔAB and ΔAC using the coordinates given for vertices A, B, and C. The area is then the absolute value of the cross product of these two vectors.
Let ΔAB be the vector from A to B and ΔAC be the vector from A to C:
- ΔAB = (Bx - Ax, By - Ay) = (3 - (-1), 2 - 5) = (4, -3)
- ΔAC = (Cx - Ax, Cy - Ay) = (0 - (-1), 2 - 5) = (1, -3)
We then take the cross product of ΔAB and ΔAC:
|Area| = | ΔAB x ΔAC |
The cross product of two-dimensional vectors is the determinant:
|Area| = | 4(-3) - (-3)(1) |
|Area| = | -12 + 3 |
|Area| = | -9 |
Thus, the area of the parallelogram is 9 square units.