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(C) Let the vertices of the equilateral triangle be $O(0, 0)$,$A(4, 0)$,and $B(x, y)$.Since the triangle is equilateral,the side length $s = \sqrt{(4-

Vedclass · Accepted Answer

$(2, \pm 2\sqrt{3})$

Vedclass · Answer

(C) Let the vertices of the equilateral triangle be $O(0, 0)$,$A(4, 0)$,and $B(x, y)$.Since the triangle is equilateral,the side length $s = \sqrt{(4-0)^2 + (0-0)^2} = 4$.The distance from $O(0, 0)$ to $B(x, y)$ must be $4$,so $x^2 + y^2 = 4^2 = 16$.The distance from $A(4, 0)$ to $B(x, y)$ must also be $4$,so $(x-4)^2 + y^2 = 4^2 = 16$.Expanding the second equation: $x^2 - 8x + 16 + y^2 = 16$.Substitute $x^2 + y^2 = 16$ into the equation: $16 - 8x + 16 = 16$,which simplifies to $8x = 16$,so $x = 2$.Substitute $x = 2$ into $x^2 + y^2 = 16$: $2^2 + y^2 = 16 \implies 4 + y^2 = 16 \implies y^2 = 12 \implies y = \pm \sqrt{12} = \pm 2\sqrt{3}$.Thus,the third vertex is $(2, \pm 2\sqrt{3})$.

If a vertex of an equilateral triangle is at the origin and the second vertex is $(4, 0)$,then its third vertex is

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