Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices

Let the given point be A (-2, -1) , B (4, 0) , C ( 3, 3) and D ( -3, 2)

So now, The slope of AB = \( \dfrac{(0+1)}{(4+2)}\)

 = \(\dfrac{1}{6}\)

The slope of CD 

= \( \dfrac{(3-2)}{(3+3)}\) = \( \dfrac{1}{6}\)

Hence, slope of AB = Slope of CD

AB ∥ CD

Now,

The slope of BC = \( \dfrac{(3-0)}{(3-4)}\)

= \( \dfrac{3}{-1}\) = -3

The slope of AD =\( \dfrac{(2+1)}{(-3+2)}\)

= \( \dfrac{3}{-1}\) = -3

Hence, slope of BC = Slope of AD

BC ∥ AD

Thus the pair of opposite sides are quadrilateral are parallel, so we can say that ABCD is a parallelogram.

Hence the given vertices, A (-2, -1), B (4, 0), C(3, 3) and D(-3, 2) are vertices of a parallelogram