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We can first draw the quadrilateral first.

$\therefore$ The area of quadrilateral $\text{PQRS} = 2 \times \text{area of } \triangle \text{PQS}$

$\qquad \qquad \qquad = 2 \times \frac{1}{2} \times \text{Base} \times \text{Height}$

$\qquad \qquad \qquad = \text{PQ} \times \text{TS} = 4 \times 2 = 8\;\text{unit}^{2}.$

Correct Answer $:\text{C}$

$\textbf{PS:}$ Consider the $xy$-plane, and suppose $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ are two points in it. Then the distance between $P_1$ and $P_2$ is $${\color{Green}{d(P_1, P_2) = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.}}$$