Question:

$AB = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.
$BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.
$CD = \sqrt{(-3 - (-1))^2 + (0 - 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$.
$DA = \sqrt{(-1 - (-3))^2 + (-2 - 0)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$. [1.5 Marks]
Diagonals:
$AC = \sqrt{(-1 - (-1))^2 + (2 - (-2))^2} = \sqrt{0 + 4^2} = 4$.
$BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} = \sqrt{(-4)^2 + 0} = 4$. [1.0 Mark]
Since all four sides are equal ($AB = BC = CD = DA = 2\sqrt{2}$) and both diagonals are equal ($AC = BD = 4$), quadrilateral $ABCD$ is a SQUARE. [0.5 Mark]

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🎯 Official CBSE Marking Scheme:
Calculating four side lengths (all equal to $\sqrt{8}$): 1.5 Marks
Calculating diagonal lengths (both equal to 4): 1.0 Mark
Concluding $ABCD$ is a Square: 0.5 Mark