One vertex of the equilateral triangle with c | vedclass

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(C) Let the coordinate of vertex $A$ be $(h, k)$. Since the triangle is equilateral,the median $AD$ is perpendicular to the side $BC$. The centroid $O

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$(-2, -2)$

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(C) Let the coordinate of vertex $A$ be $(h, k)$. Since the triangle is equilateral,the median $AD$ is perpendicular to the side $BC$. The centroid $O(0, 0)$ lies on the median $AD$. Thus,the line $OA$ is perpendicular to $BC$ $(x + y - 2 = 0)$.The slope of $BC$ is $-1$,so the slope of $OA$ must be $1$.$\frac{k - 0}{h - 0} = 1 \Rightarrow k = h$ ... $(i)$Let $D(\alpha, \beta)$ be the midpoint of $BC$. Since $O$ is the centroid,it divides the median $AD$ in the ratio $2:1$. Thus,$O = \left(\frac{2\alpha + h}{3}, \frac{2\beta + k}{3}\right) = (0, 0)$.This gives $2\alpha + h = 0 \Rightarrow \alpha = -h/2$ and $2\beta + k = 0 \Rightarrow \beta = -k/2$.Since $D(\alpha, \beta)$ lies on $x + y - 2 = 0$,we have $\alpha + \beta - 2 = 0$.Substituting $\alpha$ and $\beta$: $-h/2 - k/2 - 2 = 0 \Rightarrow h + k = -4$.Using $(i)$,$h + h = -4$ $\Rightarrow 2h = -4$ $\Rightarrow h = -2$.Thus,$k = -2$. The vertex $A$ is $(-2, -2)$.

One vertex of the equilateral triangle with centroid at the origin and one side as $x + y - 2 = 0$ is

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