If the points (1, 1),(-1, -1),and (-sqrt{3}, | vedclass

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Question

(C) Let the vertices be $A(1, 1)$,$B(-1, -1)$,and $C(-\sqrt{3}, k)$.For an equilateral triangle,the side lengths must be equal,so $AB^2 = BC^2 = AC^2$

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(C) Let the vertices be $A(1, 1)$,$B(-1, -1)$,and $C(-\sqrt{3}, k)$.For an equilateral triangle,the side lengths must be equal,so $AB^2 = BC^2 = AC^2$.First,calculate $AB^2 = (1 - (-1))^2 + (1 - (-1))^2 = 2^2 + 2^2 = 4 + 4 = 8$.Now,set $AC^2 = AB^2$:$(1 - (-\sqrt{3}))^2 + (1 - k)^2 = 8$$(1 + \sqrt{3})^2 + (1 - k)^2 = 8$$1 + 3 + 2\sqrt{3} + (1 - k)^2 = 8$$4 + 2\sqrt{3} + (1 - k)^2 = 8$$(1 - k)^2 = 4 - 2\sqrt{3}$$(1 - k)^2 = (\sqrt{3} - 1)^2$$1 - k = \pm(\sqrt{3} - 1)$.If $1 - k = \sqrt{3} - 1$,then $k = 2 - \sqrt{3}$.If $1 - k = -(\sqrt{3} - 1)$,then $1 - k = 1 - \sqrt{3}$,so $k = \sqrt{3}$.Checking $BC^2 = AB^2$ for $k = \sqrt{3}$:$BC^2 = (-1 - (-\sqrt{3}))^2 + (-1 - \sqrt{3})^2 = (\sqrt{3} - 1)^2 + (-(1 + \sqrt{3}))^2 = (3 + 1 - 2\sqrt{3}) + (1 + 3 + 2\sqrt{3}) = 4 - 2\sqrt{3} + 4 + 2\sqrt{3} = 8$.Thus,$k = \sqrt{3}$ is the correct value.

If the points $(1, 1)$,$(-1, -1)$,and $(-\sqrt{3}, k)$ are the vertices of an equilateral triangle,then what is the value of $k$?

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