Answer :

Formula used:


(3, 2), (0, 5), (–3, 2) and (0, –1)


Let the vertices be taken as A (3, 2), B (0, 5), C (–3, 2) and D (0, –1).


Distance of AB


AB = √ ((0 – 3)2 + ((5 – 2)2)


AB = √ ((–3)2 + (3)2)


AB = √ (9 + 9)


AB = √18


Distance of BC


BC= √ ((–3 – 0)2 + (2 – 5)2)


BC = √ ((–3)2 + (–3)2)


BC = √ (9 + 9)


BC = √ 18


Distance of CD


CD = √ ((0 – (–3))2 + (–1 – 2)2)


CD = √ ((0 + 3)2 + (–1 – 2)2)


CD = √ ((3)2 + (–3)2)


CD = √ (9 + 9)


CD = √18


Distance of AD


AD = √ ((0 – 3)2 + (–1 – 2)2)


AD = √ ((–3)2 + (–3)2)


AD = √ (9 + 9)


AD = √ 18


Distance of AC


AC = √ ((–3 – 3)2 + (2 – 2)2)


AC = √ ((–6)2 + (0)2)


AC = √ (36 + 0)


AC = √36


AC = 6


Distance of BD


BD = √ ((0 – 0)2 + (–1 – 5)2)


BD = √ ((0)2 + (–6)2)


BD = √ (0 + 36)


BD = √ 36


BD = 6


AB = BC = CD = DA = √18. (That is, all the sides are equal.)


AC = BD = 6. (That is, the diagonals are equal.)


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


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