Q. 9 B5.0( 1 Vote )

# Examine whether t

Formula used: (–1, 1), (0, 0), (3, 3) and (2, 4)

Let the vertices be taken as A (–1, 1), B (0, 0), C (3, 3) and D (2, 4).

Distance of AB

AB = √ ((0 – (–1))2 + (0 – 1)2)

AB = √ ((0 + 1)2 + (0 – 1)2)

AB = √ ((1)2 + (–1)2)

AB = √ (1 + 1)

AB = √ 2

Distance of BC

BC= √ ((3 – 0)2 + (3 – 0)2)

BC = √ ((3)2 + (3)2)

BC = √ (9 + 9)

BC = √ 18

Distance of CD

CD = √ ((2 – 3)2 + (4 – 3)2)

CD = √ ((1)2 + (1)2)

CD = √ (1 + 1)

CD = √2

AD = √ ((2 – (–1))2 + (4 – 1)2)

AD = √ ((2 + 1)2 + (4 – 1)2)

AD = √ ((3)2 + (3)2)

AD = √ (9 + 9)

Distance of AC

AC = √ ((3 – (–1))2 + (3 – 1)2)

AC = √ ((3 + 1)2 + (3 – 1)2)

AC = √ ((4)2 + (2)2)

AC = √ (16 + 4)

AC = √ 20

Distance of BD

BD = √ ((2 – 0)2 + (4 – 0)2)

BD = √ ((2)2 + (4)2)

BD = √ (4 + 16)

BD = √ 20

AB = CD = √2 and BC = AD = √ 18 (opposite sides of rectangle are equal).

AC = BD = √ 20 (Diagonals of rectangle are equal)

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

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