Question

Check whether the points (4,2),(2,0),(2,2) and (4,0) can be vertices of a rectangle.

Answer 1

A(4,2) B(2,0) C(2,2) D(4,0)
by the formula √(x2-x1)^2+(y2-y1)^2
AB = √(2-4)^2+(0-2)^2=√4+4=2√2
BC = √(2-2)^2+(2-0)^2=√0+4=2
CD = √(4-2)^2+(0-2)^2=√4+4=2√2
AD = √(4-4)^2+(0-2)^2=√0+4=2
since AB = CD, BC=AD
therefore ABCD is a parallelogram
AC=✓(2-4)^2+(2-2)^2=√4+0=2
BD=✓(4-2)^2+(0-0)^2=√4+0=0
since AC=BC
therefore ABCD is a rectangle