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

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(C) Let the vertices be $A(1, 1)$,$B(-1, -1)$,and $C(-\sqrt{3}, k)$.For an equilateral triangle,the distance between any two points must be equal.Firs

Vedclass · Accepted Answer

$\sqrt{3}$

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(C) Let the vertices be $A(1, 1)$,$B(-1, -1)$,and $C(-\sqrt{3}, k)$.For an equilateral triangle,the distance between any two points must be equal.First,calculate the length of side $AB$:$AB^2 = (-1 - 1)^2 + (-1 - 1)^2 = (-2)^2 + (-2)^2 = 4 + 4 = 8$.Now,calculate the distance $AC^2$:$AC^2 = (-\sqrt{3} - 1)^2 + (k - 1)^2 = (\sqrt{3} + 1)^2 + (k - 1)^2 = 3 + 1 + 2\sqrt{3} + (k - 1)^2 = 4 + 2\sqrt{3} + (k - 1)^2$.Since $AB^2 = AC^2$,we have:$8 = 4 + 2\sqrt{3} + (k - 1)^2$$(k - 1)^2 = 4 - 2\sqrt{3} = (\sqrt{3} - 1)^2$.Taking the square root,$k - 1 = \pm(\sqrt{3} - 1)$.If $k - 1 = \sqrt{3} - 1$,then $k = \sqrt{3}$.If $k - 1 = -(\sqrt{3} - 1) = 1 - \sqrt{3}$,then $k = 2 - \sqrt{3}$.Comparing with the given options,the correct value is $k = \sqrt{3}$.

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

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