# If two vertices of an equilateral triangle are (0, 0) and then find the third vertex.

Given two vertices of an equilateral traingle ABC

A (0, 0)

B (3, )

Let the third vertex be C (x, y)

We know that distance of a point A (x,y) from origin O (0, 0) is given as OA = So,    Using distance formula,   Since ∆ABC is an equilateral, AB = BC = CA.

Using CA = AB, Squaring both sides, we get

x2 + y2 = 12 …(i)

Now, using AB = BC  (Using eq (i))

Squaring both sides, we get

– 6x – 2√3y + 24 = 12

6x + 2√3y = 12

3x + √3y = 6 Substituting the value of x in eq (i), we get   12y2 – 12√3y +36 = 108

12y2 – 12√3y – 72 = 0
y2 – √3y – 6 = 0

y2 – 2√3y + √3y – 6 = 0

y (y – 2√3) + √3 (y – 2√3) = 0

(y – 2√3) (y + √3) = 0
y = 2√3 or – √3
If y = 2√3, then If y = –√3, So, the third vertex of the equilateral triangle = (0, 2√3) or (3, –√3).

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