Q. 7 C5.0( 1 Vote )

Show that the following points taken in order form the vertices of a rhombus.

(1, 0), (5, 3), (2, 7) and (−2, 4)

Answer :

Formula used:


(1, 0), (5, 3), (2, 7) and (–2, 4)


Let the vertices be taken as A (1, 0), B (5, 3), C (2, 7) and D (–2, 4).


Distance of AB


AB = ((5 1)2 + (3 0)2)


AB = √ ((4)2 + (3)2)


AB = (16 + 9)


AB = √ 25


AB = 5


Distance of BC


BC= ((2 – 5)2 + (7 3)2)


BC = √ ((3)2 + (4)2)


BC = (9 + 16)


BC = 25


BC = 5


Distance of CD


CD = ((–2 2)2 + (4 – 7)2)


CD = ((–4)2 + (–3)2)


CD = (16 + 9)


CD = 25


CD = 5


Distance of AD


AD = ((–2 – 1)2 + (4 – 0)2)


AD = ((–3)2 +(4)2)


AD = (9 + 16)


AD = √ 25


AD = 5


Distance of AC


AC = ((2 – 1)2 + (7 – 0)2)


AC = ((1)2 + (7)2)


AC = √ (1 + 49)


AC = √ 50


Distance of BD


BD = ((–2 – 5)2 + (4 3)2)


BD = ((–7)2 + (1)2)


BD = (49 + 1)


BD = 50


AB = BC = CD = DA = 10 (That is, all the sides are equal.)


AC ≠ BD (That is, the diagonals are not equal.)


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


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