If two vertices of an equilateral triangle are (3, 0) and (6, 0) find the third vertex
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Let ∆ABC be the equilateral triangle.
Let the coordinates of vertices of ∆ABC be A 3,03,0, B 6,06,0 and C (x, y).
∆ABC is an equilateral triangle.
∴ AB = BC = CA
⇒ AB2 = BC2 = CA2
⇒ [(3 – 6)2 + 0–00–02] = [(x – 6)2 + (y – 0)2] = [(x – 3)2 + (y – 0)2]
⇒ –3–32 = (x – 6)2 + y2 = (x – 3)2 + y2
⇒ 9 = (x – 6)2 + y2 = (x – 3)2 + y2
∴ (x – 6)2 + y2 = (x – 3)2 + y2
⇒ (x – 6)2 = (x – 3)2
⇒ x – 6 = ± (x – 3) BytakingsquarerootonboththesidesBytakingsquarerootonboththesides
⇒ x – 6 = x – 3 or x – 6 = –(x – 3)
⇒ x – x = –3 + 6 or x – 6 = –x + 3
⇒ 0 = 3
notpossiblenotpossible
or x + x = 3 + 6
⇒ 2x = 9

or x = 4.5

Thus, coordinates of point C are 4.5,2.64.5,2.6 or 4.5,–2.64.5,–2.6.
Let the coordinates of vertices of ∆ABC be A 3,03,0, B 6,06,0 and C (x, y).
∆ABC is an equilateral triangle.
∴ AB = BC = CA
⇒ AB2 = BC2 = CA2
⇒ [(3 – 6)2 + 0–00–02] = [(x – 6)2 + (y – 0)2] = [(x – 3)2 + (y – 0)2]
⇒ –3–32 = (x – 6)2 + y2 = (x – 3)2 + y2
⇒ 9 = (x – 6)2 + y2 = (x – 3)2 + y2
∴ (x – 6)2 + y2 = (x – 3)2 + y2
⇒ (x – 6)2 = (x – 3)2
⇒ x – 6 = ± (x – 3) BytakingsquarerootonboththesidesBytakingsquarerootonboththesides
⇒ x – 6 = x – 3 or x – 6 = –(x – 3)
⇒ x – x = –3 + 6 or x – 6 = –x + 3
⇒ 0 = 3
notpossiblenotpossible
or x + x = 3 + 6
⇒ 2x = 9

or x = 4.5

Thus, coordinates of point C are 4.5,2.64.5,2.6 or 4.5,–2.64.5,–2.6.
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