Question

Quadrilateral ABCD has vertices A(1, -3) , B(2, 1), C(6, 3), and D(7, -1). Prove or disprove that the Quad ABCD is a rectangle.

Answer 1

distance formula: \[d=\sqrt{(x2-x1)^2+(y2-y1)^2}\]
If ABCD is a rectangle, then AB=CD and AD=BC.
The length of AB is √((2-1)^2+(1+3)^2)=4.12311
The length of CD is √((7-6)^2+(-1-3)^2)=4.12311

Can you do the same for the others?