Question

If two vertices of an equilateral triangle be (0, 0), (3, `sqrt3`), find the third vertex using distance formula.

Answer 1

O(0, 0) and A(3, `sqrt3`) 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 + (sqrt3 – 0)^2 = 12,`

`OB^2 = x^2 + y^2 and, AB^2 = (x – 3)^2 + (y – sqrt3)^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 sqrt3y + 12`

`⇒ x^2 + y^2 = 12 and 6x + 2 sqrt3y = 12`

`⇒ x^2 + y^2 = 12 and 3x + sqrt3y = 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 + `sqrt3` y = 6

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

Hence, the coordinates of the third vertex B are (0, 2 `sqrt3`) or (3, – `sqrt3`).