If the two vertices of an equilateral triangle are (0, 0) and (3 = 0), find the third vertex

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).

Text Solution

Solution : Let A=`(3,sqrt(3))` and O be any point then, `O A^{2}=(3-0)^{2}+(sqrt{3}-0)^{2}=12`, `O B^{2}=x^{2}+y^{2}` and,`A B^{2}=(x-3)^{2}+(y-sqrt{3})^{2}` `Rightarrow A B^{2}=x^{2}+y^{2}-6 x-2 y+12` `therefore O A^{2}=O B^{2}=A B^{2}` `Rightarrow O A^{2}=O B^{2}` and `O B^{2}=A B^{2}` `Rightarrow x^{2}+y^{2}=12` and,`x^{2}+y^{2}=x^{2}+y^{2}-6 x-2 sqrt{3} y+12` `Rightarrow x^{2}+y^{2}=12` and `6 x+2 sqrt{3} y=12` `Rightarrow x^{2}+y^{2}=12` and `3 x+sqrt{3} y=6` `Rightarrow x^{2}+(frac{6-3 x}{sqrt{3}})^{2}=12[therefore 3 x+sqrt{3} y=6 therefore y=frac{6-3 x}{sqrt{3}}]` `Rightarrow 3 x^{2}+(6-3 x)^{2}=36` `Rightarrow 12 x^{2}-36 x=0 Rightarrow x=0,3` `therefore x=0 Rightarrow sqrt{3} y=6` `Rightarrow y=frac{6}{sqrt{3}}=2 sqrt{3}` and `x=3 Rightarrow 9+sqrt{3} y=6` `Rightarrow y=frac{6-9}{sqrt{3}}=-sqrt{3}` Hence, the coordinates of the third vertex `B` are `(0,2 sqrt{3})` or `(3,-sqrt{3})`.

Two vertices of an equilateral triangle are 0,0 and √3, √3. Find the third vertex.

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