Answer :

Formula used:


(0, –1), (2, 1), (0, 3) and (–2, 1)


Let the vertices be taken as A (0, –1), B (2, 1), C (0, 3) and D (–2, 1).


Distance of AB


AB = √ ((2 – 0)2 + ((1 – (–1))2)


AB = √ ((2 – 0))2 + (1 + 1)2)


AB = √ ((2)2 + (2)2)


AB = √ (4 + 4)


AB = √ 8


Distance of BC


BC= √ ((0 – 2)2 + (3 – 1)2)


BC = √ ((–2)2 + (2)2)


BC = √ (4 + 4)


BC = √ 8


Distance of CD


CD = √ ((–2 – 0)2 + (1 – 3)2)


CD = √ ((–2)2 + (–2)2)


CD = √ (4 + 4)


CD = √8


Distance of AD


AD = √ ((–2 – 0)2 + (1 – (–1))2)


AD = √ ((–2 – 0)2 + (1 + 1)2)


AD = √ ((–2)2 + (2)2)


AD = √ (4 + 4)


AD = √ 8


Distance of AC


AC = √ ((0 – 0)2 + (3 – (–1))2)


AC = √ ((0 – 0)2 + (3 + 1)2)


AC = √ ((0)2 + (4)2)


AC = √ (0 + 16)


AC = √ 16


AC = 4


Distance of BD


AC = √ ((–2 – 2)2 + (1 – 1)2)


AC = √ ((–4)2 + (0)2)


AC = √ (16 + 0)


AC = √ 16


AC = 4


AB = BC = CD = DA = √8 (That is, all the sides are equal.)


AC = BD = 4. (That is, the diagonals are equal.)


Hence the points A, B, C and D form a square.


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