Prove that the points A(-2, -1), B(1, 0), C(4, 3) and D(1, 2) are the vertices of a parallelogram ABCD.

Given,

A(-2, -1), B(1, 0), C(4, 3) and D(1, 2).

By using mid-point formula,

(x, y) = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

The Midpoint of Diagonal AC :

$\Rightarrow M${(AC)} = \Big(\dfrac{-2 + 4}{2}, \dfrac{-1 + 3}{2}\Big) \\[1em] \Rightarrow M{(AC)} = \Big(\dfrac{2}{2}, \dfrac{2}{2}\Big) \\[1em] \Rightarrow M_{(AC)} = (1, 1).

The Midpoint of Diagonal BD :

$\Rightarrow M${(BD)} = \Big(\dfrac{1 + 1}{2}, \dfrac{0 + 2}{2}\Big) \\[1em] \Rightarrow M{(BD)} = \Big(\dfrac{2}{2}, \dfrac{2}{2}\Big) \\[1em] \Rightarrow M_{(BD)} = (1, 1)

Since the midpoint of diagonal AC, M_AC(1, 1), is the same as the midpoint of diagonal BD, M_BD (1, 1), the diagonals AC and BD bisect each other.

We know that,

A quadrilateral is a parallelogram if its diagonals bisect each other.

Hence, proved that ABCD is a parallelogram.