The area of a square whose vertices are A(0, -2), B(3, 1), C(0, 4) and D(-3, 1) is

1. 18 sq. units

2. 15 sq. units

3. $\sqrt{18}$ units

4. $\sqrt{15}$ units

By distance formula,

$d = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} \\[1em] \therefore AB = \sqrt{(3 - 0)^2 + [1 - (-2)]^2} \\[1em] = \sqrt{3^2 + [1 + 2]^2} \\[1em] = \sqrt{9 + 3^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2}.

Area of square = (side)² = AB²

= $(3\sqrt{2})^2$ = 18 sq. units.

Hence, Option 1 is the correct option.