Answer :
Formula used:
(2, 2), (–2, –2) and (–2√3, 2√3)
Let the points be A (2, 2), B (–2, –2) and C (–2√3, 2√3)
Distance of AB
⇒ AB = √ ((–2 – 2)2 + (–2 – 2)2)
⇒ AB = √ ((–4)2 + (–4)2)
⇒ AB = √ (16 + 16)
⇒ AB = √32
⇒ AB = 4√2
Distance of BC
⇒ B C= √ ((–2√3 – (–2))2 + (2√3 – (–2))2)
⇒ B C= √ ((–2√3 + 2))2 + (2√3 + 2)2)
⇒ BC = √ (((–2√3)2 + 2 (–2√3) (2) + (2)2) + ((2√3)2 + 2 (2√3) (2) + (2)2))
⇒ BC = √ (12 – 8√3 + 4 + 12 + 8√3 + 4)
⇒ BC = √ (12 + 4 + 12 + 4
⇒ BC = √ 32
⇒ BC = 4√2
Distance of AC
⇒ AC= √ ((–2√3 – 2))2 + (2√3 – 2)2)
⇒ AC = √ (((–2√3)2 + 2 (–2√3) (–2) + (2)2) + ((2√3)2 + 2 (2√3) (–2) + (–2)2))
⇒ AC = √ (12 + 8√3 + 4 + 12 – 8√3 + 4)
⇒ AC = √ (12 + 4 + 12 + 4)
⇒ AC = √ 32
⇒ AC = 4√2
∴ AB = BC = AC = 4√2
Since, all the sides are equal the points form an equilateral triangle.
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