The points (0, 0),(a, 0),and left( frac{a}{2} | vedclass

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(B) Let $A = (0, 0)$,$B = (a, 0)$,and $C = \left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right)$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2

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Equilateral triangle

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(B) Let $A = (0, 0)$,$B = (a, 0)$,and $C = \left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right)$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB = \sqrt{(a - 0)^2 + (0 - 0)^2} = \sqrt{a^2} = a$.$BC = \sqrt{\left( \frac{a}{2} - a \right)^2 + \left( \frac{a\sqrt{3}}{2} - 0 \right)^2} = \sqrt{\left( -\frac{a}{2} \right)^2 + \left( \frac{a\sqrt{3}}{2} \right)^2} = \sqrt{\frac{a^2}{4} + \frac{3a^2}{4}} = \sqrt{a^2} = a$.$AC = \sqrt{\left( \frac{a}{2} - 0 \right)^2 + \left( \frac{a\sqrt{3}}{2} - 0 \right)^2} = \sqrt{\frac{a^2}{4} + \frac{3a^2}{4}} = \sqrt{a^2} = a$.Since $AB = BC = AC = a$,the triangle is an equilateral triangle.

The points $(0, 0)$,$(a, 0)$,and $\left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right)$ are vertices of:

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