To plot the points, we mark $A(2,\, 1)$, $B(-1,\, 2)$, $C(-2,\, -1)$, and $D(1,\, -2)$ on the coordinate plane. Point $A$ lies in Quadrant I, $B$ in Quadrant II, $C$ in Quadrant III, and $D$ in Quadrant IV.

Next, we find the lengths of the sides using the Baudhāyana–Pythagoras Theorem. The distance between two points is the square root of the sum of the squares of the distances moved along the x-axis and y-axis.

$$\textbf{AB} = \sqrt{[(-1 - 2)^2 + (2 - 1)^2]} = \sqrt{(9 + 1)} = \sqrt{10} \text{ units.}$$ $$\textbf{BC} = \sqrt{[(-2 - (-1))^2 + (-1 - 2)^2]} = \sqrt{(1 + 9)} = \sqrt{10} \text{ units.}$$ $$\textbf{CD} = \sqrt{[(1 - (-2))^2 + (-2 - (-1))^2]} = \sqrt{(9 + 1)} = \sqrt{10} \text{ units.}$$ $$\textbf{DA} = \sqrt{[(2 - 1)^2 + (1 - (-2))^2]} = \sqrt{(1 + 9)} = \sqrt{10} \text{ units.}$$

We also find the lengths of the diagonals AC and BD.

$$\textbf{AC} = \sqrt{[(-2 - 2)^2 + (-1 - 1)^2]} = \sqrt{(16 + 4)} = \sqrt{20} \text{ units.}$$ $$\textbf{BD} = \sqrt{[(1 - (-1))^2 + (-2 - 2)^2]} = \sqrt{(4 + 16)} = \sqrt{20} \text{ units.}$$

Since all four sides are equal ($AB = BC = CD = DA = \sqrt{10}$) and both diagonals are equal ($AC = BD = \sqrt{20}$), the quadrilateral $ABCD$ is a square.

Finally, the area of this square is calculated as $(\text{side})^2$.

$$\textbf{Area} = (\sqrt{10})^2 = \mathbf{10 \text{ square units.}}$$