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)

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

AD = ((7 15))2 + (2 – 20)2)

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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