# If Two Vertices of an Equilateral Triangle Be (0, 0), (3, √3 ), Find the Third Vertex - Mathematics

Sum

If two vertices of an equilateral triangle be (0, 0), (3, √3 ), find the third vertex

#### Solution

O(0, 0) and A(3, √3) be the given points and let B(x, y) be the third vertex of equilateral ∆OAB. Then, OA = OB = AB

⇒ OA^2 = OB^2 = AB^2

We have,

OA^2 = (3 – 0)^2 + (√3 – 0)^2 = 12,

OB^2 = x^2 + y^2 and, AB^2 = (x – 3)^2 + (y – √3)^2

⇒ AB^2 = x^2 + y^2 – 6x – 2 y + 12

∴ OA^2 = OB^2 = AB^2

⇒ OA^2 = OB^2 and OB^2 = AB^2

⇒ x^2 + y^2 = 12

and, x^2 + y^2 = x^2 + y^2 – 6x – 2 √3y + 12

⇒ x^2 + y^2 = 12 and 6x + 2 √3y = 12

⇒ x^2 + y^2 = 12 and 3x + √3y = 6

=>x^2+((6-3x)/sqrt3)^2=12[.:3x+sqrt3y=6:.y=(6-3x)/sqrt3]

⇒ 3x^2 + (6 – 3x)^2 = 36

⇒ 12x^2 – 36x = 0 ⇒ x = 0, 3

∴ x = 0 ⇒ √3y = 6

\Rightarrow y=\frac{6}{\sqrt{3}}=2\sqrt{3}[

and, x = 3 ⇒ 9 + √3 y = 6

=>y=(6-9)/sqrt3=-sqrt3

Hence, the coordinates of the third vertex B are (0, 2 √3) or (3, – √3).

Concept: Distance Formula
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