The three sides of a triangle are given by (x | vedclass

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(C) The equation of the sides is given by $(x^2+7xy+2y^2)(y-1)=0$. This implies the sides are $x^2+7xy+2y^2=0$ and $y-1=0$. However,a triangle is form

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$(\frac{-7}{3}, \frac{2}{3})$

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(C) The equation of the sides is given by $(x^2+7xy+2y^2)(y-1)=0$. This implies the sides are $x^2+7xy+2y^2=0$ and $y-1=0$. However,a triangle is formed by three straight lines. The equation $x^2+7xy+2y^2=0$ represents two lines passing through the origin $(0,0)$. Let these lines be $L_1$ and $L_2$. The third line is $L_3: y=1$. To find the vertices,we find the intersection points: $1$. Intersection of $L_1$ and $L_2$: Both pass through the origin,so one vertex is $V_1 = (0,0)$. $2$. Intersection of $L_1$ and $L_3$: Substitute $y=1$ into $x^2+7xy+2y^2=0$,we get $x^2+7x+2=0$. Let the roots be $x_1, x_2$. The vertices are $V_2 = (x_1, 1)$ and $V_3 = (x_2, 1)$. Using the quadratic formula for $x^2+7x+2=0$,the sum of roots $x_1+x_2 = -7$. The centroid $(G_x, G_y)$ is given by $(\frac{0+x_1+x_2}{3}, \frac{0+1+1}{3}) = (\frac{-7}{3}, \frac{2}{3})$.

The three sides of a triangle are given by $(x^2+7xy+2y^2)(y-1)=0$. Then the centroid of that triangle is

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