7. The three vertices of a parallelogram ABCD are \( \mathrm{A}(-1,-2), \mathrm{B}(4,-1) \) and C \( (6,3) \). Find the coordinates of vertex D and find the area of parallelogram \( \mathbf{A B C D} \).

To find the coordinates of vertex D and the area of parallelogram ABCD, we can use the properties of a parallelogram. Given: - Vertex A: (-1, -2) - Vertex B: (4, -1) - Vertex C: (6, 3) Since ABCD is a parallelogram, the diagonals bisect each other. This means that the midpoint of diagonal AC will be the same as the midpoint of diagonal BD. Let's find the midpoint of diagonal AC: Midpoint of AC = ((-1 + 6)/2, (-2 + 3)/2) = (5/2, 1/2) Now, we need to find the coordinates of vertex D such that the midpoint of BD is (5/2, 1/2). Let the coordinates of vertex D be (x, y). Then, the midpoint of BD is ((4 + x)/2, (-1 + y)/2). Since the midpoint of BD is (5/2, 1/2), we can set up the following equations: ((4 + x)/2) = 5/2 (-1 + y)/2 = 1/2 Solving these equations will give us the coordinates of vertex D. Once we have the coordinates of vertex D, we can find the area of parallelogram ABCD using the formula for the area of a parallelogram: Area = |(x1 * y2 - x2 * y1)| where (x1, y1) and (x2, y2) are the coordinates of two adjacent vertices. Let's proceed with finding the coordinates of vertex D and then the area of parallelogram ABCD. Solve the system of equations by following steps: - step0: Solution: \(\left\{ \begin{array}{l}\frac{\left(4+x\right)}{2}=\frac{5}{2}\\\frac{\left(-1+y\right)}{2}=\frac{1}{2}\end{array}\right.\) - step1: Remove the parentheses: \(\left\{ \begin{array}{l}\frac{4+x}{2}=\frac{5}{2}\\\frac{-1+y}{2}=\frac{1}{2}\end{array}\right.\) - step2: Calculate: \(\left\{ \begin{array}{l}x=1\\y=2\end{array}\right.\) - step3: Check the solution: \(\left\{ \begin{array}{l}x=1\\y=2\end{array}\right.\) - step4: Rewrite: \(\left(x,y\right) = \left(1,2\right)\) The coordinates of vertex D are (1, 2). Now, let's find the area of parallelogram ABCD using the formula: Area = |(x1 * y2 - x2 * y1)| where (x1, y1) = (-1, -2) and (x2, y2) = (4, -1). Area = |(-1 * -1 - 4 * -2)| = |(1 + 8)| = |9| = 9 Therefore, the area of parallelogram ABCD is 9 square units.