The points left( frac{a}{sqrt{3}}, a right),l | vedclass

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(B) Let $A = \left( \frac{a}{\sqrt{3}}, a \right)$,$B = \left( \frac{2a}{\sqrt{3}}, 2a \right)$,and $C = \left( \frac{a}{\sqrt{3}}, 3a \right)$.Calcul

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An isosceles triangle

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(B) Let $A = \left( \frac{a}{\sqrt{3}}, a \right)$,$B = \left( \frac{2a}{\sqrt{3}}, 2a \right)$,and $C = \left( \frac{a}{\sqrt{3}}, 3a \right)$.Calculate the squared distances between the vertices:$AB^2 = \left( \frac{a}{\sqrt{3}} - \frac{2a}{\sqrt{3}} \right)^2 + (a - 2a)^2 = \left( -\frac{a}{\sqrt{3}} \right)^2 + (-a)^2 = \frac{a^2}{3} + a^2 = \frac{4a^2}{3}$.$BC^2 = \left( \frac{2a}{\sqrt{3}} - \frac{a}{\sqrt{3}} \right)^2 + (2a - 3a)^2 = \left( \frac{a}{\sqrt{3}} \right)^2 + (-a)^2 = \frac{a^2}{3} + a^2 = \frac{4a^2}{3}$.$AC^2 = \left( \frac{a}{\sqrt{3}} - \frac{a}{\sqrt{3}} \right)^2 + (a - 3a)^2 = 0^2 + (-2a)^2 = 4a^2$.Since $AB^2 = BC^2 = \frac{4a^2}{3}$,the sides $AB$ and $BC$ are equal in length.Therefore,the triangle is an isosceles triangle.

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

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