Let A(–2, –1), B(4, 0), C(3, 3) and D(–3, 2) be the vertices of the given quadrilateral ABCD. Then,

Slope of AB =

0−(−1)

4−(−2)

=

1

6

Slope of DC =

3−2

3−(−3)

=

1

6

Slope of BC =

3−0

3−4

=

3

−1

= - 3
Slope of AD =

2−(−1)

−3−(−2)

=

3

−1

= - 3
Slope of AD = slope of BC ⇒ AD || BC
Slope of AB = Slope of CD ⇒ AB || CD
Hence, ABCD is a parallelogram.