If the points (0, 0),(2, 2sqrt{3}) and (a, b) | vedclass

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(C) Let the vertices be $O(0, 0)$,$A(2, 2\sqrt{3})$,and $B(a, b)$.Since it is an equilateral triangle,the side length $l$ is the distance $OA = \sqrt{

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$(4, 0)$

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(C) Let the vertices be $O(0, 0)$,$A(2, 2\sqrt{3})$,and $B(a, b)$.Since it is an equilateral triangle,the side length $l$ is the distance $OA = \sqrt{(2-0)^2 + (2\sqrt{3}-0)^2} = \sqrt{4 + 12} = \sqrt{16} = 4$.Thus,$OB^2 = a^2 + b^2 = 4^2 = 16$ (Equation $1$).Also,$AB^2 = (a-2)^2 + (b-2\sqrt{3})^2 = 16$.Expanding this: $a^2 - 4a + 4 + b^2 - 4\sqrt{3}b + 12 = 16$.Substituting $a^2 + b^2 = 16$: $16 - 4a - 4\sqrt{3}b + 16 = 16$,which simplifies to $4a + 4\sqrt{3}b = 16$,or $a + \sqrt{3}b = 4$ (Equation $2$).From Equation $2$,$a = 4 - \sqrt{3}b$. Substituting into Equation $1$: $(4 - \sqrt{3}b)^2 + b^2 = 16$.$16 - 8\sqrt{3}b + 3b^2 + b^2 = 16 \Rightarrow 4b^2 - 8\sqrt{3}b = 0$.$4b(b - 2\sqrt{3}) = 0$,so $b = 0$ or $b = 2\sqrt{3}$.If $b = 0$,$a = 4$. If $b = 2\sqrt{3}$,$a = 4 - 6 = -2$.Since $(a, b)$ must be distinct from $(2, 2\sqrt{3})$,the solution is $(4, 0)$.

If the points $(0, 0)$,$(2, 2\sqrt{3})$ and $(a, b)$ are the vertices of an equilateral triangle,then $(a, b) = $

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