Q. 8 A5.0( 1 Vote )

# Examine whether t

Formula used:

(0, –1), (2, 1), (0, 3) and (–2, 1)

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

Distance of AB

AB = √ ((2 – 0)2 + ((1 – (–1))2)

AB = √ ((2 – 0))2 + (1 + 1)2)

AB = √ ((2)2 + (2)2)

AB = √ (4 + 4)

AB = √ 8

Distance of BC

BC= √ ((0 – 2)2 + (3 – 1)2)

BC = √ ((–2)2 + (2)2)

BC = √ (4 + 4)

BC = √ 8

Distance of CD

CD = √ ((–2 – 0)2 + (1 – 3)2)

CD = √ ((–2)2 + (–2)2)

CD = √ (4 + 4)

CD = √8

AD = √ ((–2 – 0)2 + (1 – (–1))2)

AD = √ ((–2 – 0)2 + (1 + 1)2)

AD = √ ((–2)2 + (2)2)

AD = √ (4 + 4)

Distance of AC

AC = √ ((0 – 0)2 + (3 – (–1))2)

AC = √ ((0 – 0)2 + (3 + 1)2)

AC = √ ((0)2 + (4)2)

AC = √ (0 + 16)

AC = √ 16

AC = 4

Distance of BD

AC = √ ((–2 – 2)2 + (1 – 1)2)

AC = √ ((–4)2 + (0)2)

AC = √ (16 + 0)

AC = √ 16

AC = 4

AB = BC = CD = DA = √8 (That is, all the sides are equal.)

AC = BD = 4. (That is, the diagonals are equal.)

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

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