In Delta ABC,the coordinates of B are (0, 0), | vedclass

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(B) Given $B = (0, 0)$ and the midpoint of $BC$ is $(2, 0)$.Let $C = (h, k)$. By the midpoint formula,$\frac{h+0}{2} = 2 \implies h = 4$ and $\frac{k+

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$\left(\frac{5}{3}, \frac{1}{\sqrt{3}}\right)$

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(B) Given $B = (0, 0)$ and the midpoint of $BC$ is $(2, 0)$.Let $C = (h, k)$. By the midpoint formula,$\frac{h+0}{2} = 2 \implies h = 4$ and $\frac{k+0}{2} = 0 \implies k = 0$. Thus,$C = (4, 0)$.Since $AB = 2$ and $\angle ABC = 60^{\circ}$,the coordinates of $A$ are $(2 \cos 60^{\circ}, 2 \sin 60^{\circ}) = (2 	imes \frac{1}{2}, 2 	imes \frac{\sqrt{3}}{2}) = (1, \sqrt{3})$.The centroid $G$ of $\Delta ABC$ with vertices $A(1, \sqrt{3})$,$B(0, 0)$,and $C(4, 0)$ is given by:$G = \left(\frac{1+0+4}{3}, \frac{\sqrt{3}+0+0}{3}ight) = \left(\frac{5}{3}, \frac{\sqrt{3}}{3}ight) = \left(\frac{5}{3}, \frac{1}{\sqrt{3}}ight)$.

In $\Delta ABC$,the coordinates of $B$ are $(0, 0)$,$AB = 2$,$\angle ABC = \frac{\pi}{3}$,and the coordinates of the midpoint of $BC$ are $(2, 0)$. Find the centroid of the triangle.

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