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Four points $\text{P(0, 1), Q(0, – 3), R( – 2, – 1),}$ and $\text{S(2, – 1)}$ represent the vertices of a quadrilateral.

What is the area enclosed by the quadrilateral?

  1. $4$
  2. $4 \sqrt{2}$
  3. $8$
  4. $8 \sqrt{2}$
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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 $${\color{Green}{d(P_1, P_2) = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.}}$$

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Given points: \( P(0, 1) \), \( Q(0, -3) \), \( R(-2, -1) \), and \( S(2, -1) \).

The height (\( h \)) is the absolute difference in the y-coordinates of \( P \) and \( Q \):
\[ h = |1 - (-3)| = 4 \]

The base (\( b \)) is the absolute difference in the x-coordinates of \( R \) and \( S \):
\[ b = |-2 - 2| = 4 \]

The area (\( A \)) of the quadrilateral is given by:
\[ A = \frac{1}{2} \times b \times h \]

Substituting the values:
\[ A = \frac{1}{2} \times 4 \times 4 = 8 \]

Therefore, the area of the quadrilateral is \( 8 \).
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