Answer :

Formula used:


(–1, 1), (0, 0), (3, 3) and (2, 4)


Let the vertices be taken as A (–1, 1), B (0, 0), C (3, 3) and D (2, 4).


Distance of AB


AB = √ ((0 – (–1))2 + (0 – 1)2)


AB = √ ((0 + 1)2 + (0 – 1)2)


AB = √ ((1)2 + (–1)2)


AB = √ (1 + 1)


AB = √ 2


Distance of BC


BC= √ ((3 – 0)2 + (3 – 0)2)


BC = √ ((3)2 + (3)2)


BC = √ (9 + 9)


BC = √ 18


Distance of CD


CD = √ ((2 – 3)2 + (4 – 3)2)


CD = √ ((1)2 + (1)2)


CD = √ (1 + 1)


CD = √2


Distance of AD


AD = √ ((2 – (–1))2 + (4 – 1)2)


AD = √ ((2 + 1)2 + (4 – 1)2)


AD = √ ((3)2 + (3)2)


AD = √ (9 + 9)


AD = √ 18


Distance of AC


AC = √ ((3 – (–1))2 + (3 – 1)2)


AC = √ ((3 + 1)2 + (3 – 1)2)


AC = √ ((4)2 + (2)2)


AC = √ (16 + 4)


AC = √ 20


Distance of BD


BD = √ ((2 – 0)2 + (4 – 0)2)


BD = √ ((2)2 + (4)2)


BD = √ (4 + 16)


BD = √ 20


AB = CD = √2 and BC = AD = √ 18 (opposite sides of rectangle are equal).


AC = BD = √ 20 (Diagonals of rectangle are equal)


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


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