Answer:
[tex]\left\begin{array}{ccc}A\left( \boxed{3} , \boxed{8} \right)&B\left( \boxed{9} , \boxed{6} \right )\\\\C\left( \boxed{1} , \boxed{4} \right)&D\left( \boxed{7} , \boxed{2} \right )\end{array}\right[/tex]
Step-by-step explanation:
The coordinates are:
[tex]\left\begin{array}{ccc}A\left( \boxed{3} , \boxed{8} \right)&B\left( \boxed{9} , \boxed{6} \right )\\\\C\left( \boxed{1} , \boxed{4} \right)&D\left( \boxed{7} , \boxed{2} \right )\end{array}\right[/tex]
The parallelogram is attached below.
To verify if these coordinates form a parallelogram, we show that:
Using Distance Formula
[tex]AB = \sqrt{(6-8)^2+(9-3)^2} = \sqrt{(-2)^2+(6)^2} = \sqrt{40} $ units[/tex]
[tex]CD = \sqrt{(2-4)^2+(7-1)^2} = \sqrt{(-2)^2+(6)^2} = \sqrt{40} $ units[/tex]
[tex]AC = \sqrt{(8-4)^2+(3-1)^2} = \sqrt{(4)^2+(2)^2} = \sqrt{20} $ units[/tex]
[tex]BD = \sqrt{(6-2)^2+(9-7)^2} = \sqrt{(4)^2+(2)^2} = \sqrt{20} $ units[/tex]
Since AB=CD; and AC=BD, the coordinates A, B, C, and D form the vertex of a parallelogram.
