Plot the points $\mathrm{A}(2, 1), \mathrm{B}(-1, 2), \mathrm{C}(-2, -1),$ and $\mathrm{D}(1, -2)$ in the coordinate plane. Is $\mathrm{ABCD}$ a square? Can you explain why? What is the area of this square?

Plotting of points A (2, 1), B (-1, 2), C (-2, -1), and D (1, -2) in the coordinate plane is given as:

A square is a quadrilateral whose four sides are equal and whose adjacent sides are perpendicular.
Using distance formula,

$$AB = \sqrt{(-1 - 2)^2 + (2 - 1)^2} = \sqrt{9 + 1} = \sqrt{10}$$

$$BC = \sqrt{(-2 - (-1))^2 + (-1 - 2)^2} = \sqrt{1 + 9} = \sqrt{10}$$

$$CD = \sqrt{(1 - (-2))^2 + (-2 - (-1))^2} = \sqrt{9 + 1} = \sqrt{10}$$

$$DA = \sqrt{(2 - 1)^2 + (1 - (-2))^2} = \sqrt{1 + 9} = \sqrt{10}$$

All four sides equal $\sqrt{10}$ , so ABCD is at least a rhombus.

$$AC = \sqrt{(-2 - 2)^2 + (-1 - 1)^2} = \sqrt{16 + 4} = \sqrt{20}$$

$$BD = \sqrt{(1 - (-1))^2 + (-2 - 2)^2} = \sqrt{4 + 16} = \sqrt{20}$$

Since, all sides and diagonals are equal, therefore, it is a square.

Area of square = $(\text{side})^2 = (\sqrt{10})^2 = 10 \text{ sq. units}$ .