Q. 7 E5.0( 1 Vote )

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

(15, 20), (−3, 12), (−11, −6) and (7, 2)

Answer :

Formula used:


(15, 20), (–3, 12), (–11, –6) and (7, 2)


Let the vertices be taken as A (15, 20), B (–3, 12), C (–11, –6) and D (7, 2).


Distance of AB


AB = ((–3 – 15)2 + (12 20)2)


AB = ((–18)2 + (–8)2)


AB = √ (324 + 64)


AB = 388


Distance of BC


BC= ((–11 –(–3))2 + (–6 12)2)


BC = √ (–11 + 3)2 + (–6 – 12)2)


BC = √ ((–8)2 + (–18)2)


BC = (64 + 324)


BC = 388


Distance of CD


CD = ((7 – (–11))2 + (2 – (–6))2)


CD = ((7 + 11)2 + (2 + 6)2)


CD = ((18)2 + (8)2)


CD = (324 + 64)


CD = 388


Distance of AD


AD = ((7 15))2 + (2 – 20)2)


AD = ((–8)2 +(–18)2)


AD = (64 + 324)


AD = √ 388


Distance of AC


AC = ((–11 15)2 + (–6 – 20)2)


AC = ((–26)2 +(–26)2)


AC = √ (676 + 676)


AC = √ 1352


Distance of BD


BD = ((7 – (–3))2 + (2 12)2)


BD = ((7 + 3))2 + (2 12)2)


BD = ((10)2 + (–10)2)


BD = (100 + 100)


BD = 200


AB = BC = CD = DA = √388 (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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