Q. 8 C5.0( 1 Vote )

# Examine whether t

Answer :

Formula used: (3, 2), (0, 5), (–3, 2) and (0, –1)

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

Distance of AB

AB = √ ((0 – 3)2 + ((5 – 2)2)

AB = √ ((–3)2 + (3)2)

AB = √ (9 + 9)

AB = √18

Distance of BC

BC= √ ((–3 – 0)2 + (2 – 5)2)

BC = √ ((–3)2 + (–3)2)

BC = √ (9 + 9)

BC = √ 18

Distance of CD

CD = √ ((0 – (–3))2 + (–1 – 2)2)

CD = √ ((0 + 3)2 + (–1 – 2)2)

CD = √ ((3)2 + (–3)2)

CD = √ (9 + 9)

CD = √18

Distance of AD

AD = √ ((0 – 3)2 + (–1 – 2)2)

AD = √ ((–3)2 + (–3)2)

AD = √ (9 + 9)

AD = √ 18

Distance of AC

AC = √ ((–3 – 3)2 + (2 – 2)2)

AC = √ ((–6)2 + (0)2)

AC = √ (36 + 0)

AC = √36

AC = 6

Distance of BD

BD = √ ((0 – 0)2 + (–1 – 5)2)

BD = √ ((0)2 + (–6)2)

BD = √ (0 + 36)

BD = √ 36

BD = 6

AB = BC = CD = DA = √18. (That is, all the sides are equal.)

AC = BD = 6. (That is, the diagonals are equal.)

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

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