Leave the given point alone \[\mathbf{A}\text{ }\left( -\text{ }\mathbf{2},\text{ }-\text{ }\mathbf{1} \right)\text{ },\text{ }\mathbf{B}\text{ }\left( \mathbf{4},\text{ }\mathbf{0} \right)\text{ },\text{ }\mathbf{C}\text{ }\left( \text{ }\mathbf{3},\text{ }\mathbf{3} \right)\text{ }\mathbf{and}\text{ }\mathbf{D}\text{ }\left( \text{ }-\text{ }\mathbf{3},\text{ }\mathbf{2} \right)\]

So presently, The incline of \[\mathbf{AB}\text{ }=\text{ }\left( \mathbf{0}+\mathbf{1} \right)/\left( \mathbf{4}+\mathbf{2} \right)\text{ }=\text{ }\mathbf{1}/\mathbf{6}\]

The incline of \[\mathbf{CD}\text{ }=\text{ }\left( \mathbf{3}-\mathbf{2} \right)/\left( \mathbf{3}+\mathbf{3} \right)\text{ }=\text{ }\mathbf{1}/\mathbf{6}\]

Consequently, slant of AB = Slope of CD

\[\therefore \mathbf{AB}\parallel \mathbf{CD}\]

Presently,

The incline of \[\mathbf{BC}\text{ }=\text{ }\left( \mathbf{3}-\mathbf{0} \right)/\left( \mathbf{3}-\mathbf{4} \right)\text{ }=\text{ }\mathbf{3}/\text{ }-\text{ }\mathbf{1}\text{ }=\text{ }-\text{ }\mathbf{3}\]

The incline of \[\mathbf{AD}\text{ }=\text{ }\left( \mathbf{2}+\mathbf{1} \right)/\left( -\text{ }\mathbf{3}+\mathbf{2} \right)\text{ }=\text{ }\mathbf{3}/\text{ }-\text{ }\mathbf{1}\text{ }=\text{ }-\text{ }\mathbf{3}\]

Consequently, slant of BC = Slope of AD

\[\therefore \mathbf{BC}\parallel \mathbf{AD}\]

Accordingly the pair of inverse sides are quadrilateral are equal, so we can say that ABCD is a parallelogram.

Consequently the given vertices, \[\mathbf{A}\text{ }\left( -\text{ }\mathbf{2},\text{ }-\text{ }\mathbf{1} \right),\text{ }\mathbf{B}\text{ }\left( \mathbf{4},\text{ }\mathbf{0} \right),\text{ }\mathbf{C}\left( \mathbf{3},\text{ }\mathbf{3} \right)\text{ }\mathbf{and}\text{ }\mathbf{D}\left( -\text{ }\mathbf{3},\text{ }\mathbf{2} \right)\] are vertices of a parallelogram.