If A (0,0) and B (1, sqrt{3}) are the vertice | vedclass

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(C) Let the third vertex be $C(x, y)$. Since $ABC$ is an equilateral triangle,$AB = BC = AC$.The distance $AB = \sqrt{(1-0)^2 + (\sqrt{3}-0)^2} = \sqr

Vedclass · Accepted Answer

Both $(2, 0)$ and $(-1, \sqrt{3})$

Vedclass · Answer

(C) Let the third vertex be $C(x, y)$. Since $ABC$ is an equilateral triangle,$AB = BC = AC$.The distance $AB = \sqrt{(1-0)^2 + (\sqrt{3}-0)^2} = \sqrt{1 + 3} = 2$.Since $AB = BC = AC = 2$,we have:$x^2 + y^2 = 2^2 = 4$  ---$(1)$$(x-1)^2 + (y-\sqrt{3})^2 = 4$  ---$(2)$Expanding $(2)$: $x^2 - 2x + 1 + y^2 - 2\sqrt{3}y + 3 = 4$.Substituting $x^2 + y^2 = 4$ into the equation: $4 - 2x + 1 - 2\sqrt{3}y + 3 = 4$,which simplifies to $2x + 2\sqrt{3}y = 4$ or $x + \sqrt{3}y = 2$.Thus,$x = 2 - \sqrt{3}y$.Substituting this into $(1)$: $(2 - \sqrt{3}y)^2 + y^2 = 4$.$4 - 4\sqrt{3}y + 3y^2 + y^2 = 4$,so $4y^2 - 4\sqrt{3}y = 0$.$4y(y - \sqrt{3}) = 0$,giving $y = 0$ or $y = \sqrt{3}$.If $y = 0$,$x = 2 - 0 = 2$. If $y = \sqrt{3}$,$x = 2 - \sqrt{3}(\sqrt{3}) = 2 - 3 = -1$.Therefore,the third vertex $C$ can be $(2, 0)$ or $(-1, \sqrt{3})$.

If $A (0,0)$ and $B (1, \sqrt{3})$ are the vertices of an equilateral triangle,then find the third vertex.

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