Answer:
Given the vertices of the rectangle ABCD:
A = (1, 7) , B = (5, 3), C = (3,1) and D = (-1, 5)
Distance(D) formula for two points is given by;
[tex]D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using distance formula:
[tex]AB = \sqrt{(5-1)^2+(3-7)^2}=\sqrt{(4)^2+(-4)^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}[/tex] units
[tex]BC= \sqrt{(3-5)^2+(1-3)^2}=\sqrt{(-2)^2+(-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}[/tex] units.
[tex]CD = \sqrt{(-1-3)^2+(5-1)^2}=\sqrt{(-4)^2+(4)^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}[/tex] units
[tex]DA= \sqrt{(-1-1)^2+(7-5)^2}=\sqrt{(-2)^2+(2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}[/tex] units.
Since, the Opposite sides of a rectangle are the same length.
⇒AB = CD and BC =DA
Area of rectangle is equal to multiply its width by length.
Area of rectangle ABCD = [tex]CD \times BC[/tex]
= [tex]4\sqrt{2} \times 2\sqrt{2} = 8 \times 2 = 16[/tex] square units.
Therefore, the area of rectangle is, 16 square units.
