A quadrilateral has vertices in the order (0, -1), (6, 7), (-2,3) and…

Let's start naming these points ,

A( 0, -1), B(6 , 7) , C ( -2, 3 ) and D ( 8 , 3 )

Now,

AD = \(\sqrt { ( 8 - 0)² + ( 3 + 1 )² }\)

= \(\sqrt { 64+ 16 }\)

= \(\sqrt {80 }\)

= 4\(\sqrt { 5 }\)

BC = \(\sqrt { ( - 2 - 6 )² + ( 3 - 7  )² }\)

= \(\sqrt { 64+ 16 }\)

= \(\sqrt {80 }\)

= 4\(\sqrt { 5 }\)

AC = \(\sqrt { ( -2 - 0)² + ( 3 + 1 )² }\)

= \(\sqrt { 4+ 16 }\)

= \(\sqrt {20 }\)

= 2\(\sqrt { 5 }\)

And BD = \(\sqrt { ( 8- 6)² + (3 - 7  )² }\)

= \(\sqrt { 4 + 16 }\)

= \(\sqrt {20 }\)

= 2\(\sqrt { 5 }\)

That means , AD = BC and AC = BD

Hence , ABCD is a parallelogram .

Now,

AB = \(\sqrt { ( 6 - 0)² + ( 7 + 1 )² }\)

= \(\sqrt { 36 + 64 }\)

= \(\sqrt {100 }\)

= 10

CD =\(\sqrt { (8 + 2 )² + ( 3 - 3 )² }\)

= \(\sqrt { 100 + 0  }\)

= 10

That means , AB = CD

Now,

AB² = AD² + DB²  

10² = ( 4\(\sqrt { 5 }\) ) ² + ( 2\(\sqrt { 5 }\) ) ²

100 = 80 + 20

100 = 100    ( followed )

Similarly ,

CD² = CB² + BD²  

10² = ( 4\(\sqrt { 5 }\) ) ² + ( 2\(\sqrt { 5 }\) ) ²

100 = 80 + 20

100 = 100    ( followed )

Hence it is concluded that :- ABCD is a rectangle.