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)

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

AD = ((–2 – 1)2 + (4 – 0)2)

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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