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).