Questions: Find the area of the parallelogram whose vertices are given A(-2,2) B(2,0) C(9,1) D(5,3)

Solution Steps

To find the area of a parallelogram given its vertices, we can use the determinant method. The area can be calculated using the coordinates of the vertices. We can choose any two adjacent sides of the parallelogram, find the vectors representing these sides, and then compute the cross product of these vectors. The magnitude of the cross product gives the area of the parallelogram.

Given the vertices of the parallelogram: \[ A(-2, 2), \quad B(2, 0), \quad C(9, 1), \quad D(5, 3) \] We choose vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AD}\): \[ \overrightarrow{AB} = B - A = [2 - (-2), 0 - 2] = [4, -2] \] \[ \overrightarrow{AD} = D - A = [5 - (-2), 3 - 2] = [7, 1] \]

The cross product of two vectors \(\overrightarrow{u} = [u_1, u_2, 0]\) and \(\overrightarrow{v} = [v_1, v_2, 0]\) in 3D space is given by: \[ \overrightarrow{u} \times \overrightarrow{v} = [0, 0, u_1 v_2 - u_2 v_1] \] Applying this to our vectors: \[ \overrightarrow{AB} \times \overrightarrow{AD} = [0, 0, 4 \cdot 1 - (-2) \cdot 7] = [0, 0, 4 + 14] = [0, 0, 18] \]

The magnitude of the cross product vector \([0, 0, 18]\) is: \[ \|\overrightarrow{AB} \times \overrightarrow{AD}\| = \sqrt{0^2 + 0^2 + 18^2} = \sqrt{324} = 18 \]

Final Answer