What is the area of a parallelogram whose vertices are A(−4, 9) , B(11, 9) , C(5, −1) , and D(−10, −1) ?

Let [tex]A(-4,9)\\B(11,9)\\C(5,-1) \\D(-10,-1)\\E(-4.-1)[/tex]  using a graphing tool  see the attached figure to better understand the problem we know that Parallelogram is a quadrilateral with opposite sides parallel and equal in length so [tex]AB=CD \\AD=BC[/tex] The area of a parallelogram is equal to [tex]A=B*h[/tex]  where  B is the base h is the height  the base B is equal to the distance AB the height h is equal to the distance AE  Step 1 Find the distance AB the formula to calculate the distance between two points is equal to [tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex] [tex]A(-4,9)\\B(11,9)[/tex]  substitute the values [tex]d=\sqrt{(9-9)^{2}+(11+4)^{2}}[/tex] [tex]d=\sqrt{(0)^{2}+(15)^{2}}[/tex] [tex]dAB=15\ units[/tex] Step 2 Find the distance AE the formula to calculate the distance between two points is equal to [tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex] [tex]A(-4,9)\\E(-4.-1)[/tex]  substitute the values [tex]d=\sqrt{(-1-9)^{2}+(-4+4)^{2}}[/tex] [tex]d=\sqrt{(-10)^{2}+(0)^{2}}[/tex] [tex]dAE=10\ units[/tex] Step 3 Find the area of the parallelogram The area of a parallelogram is equal to [tex]A=B*h[/tex] [tex]A=AB*AE[/tex] substitute the values [tex]A=15*10=150\ units^{2}[/tex] therefore the answer is the area of the parallelogram is [tex]150\ units^{2}[/tex]