The centroid of the triangle formed by the li | vedclass

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(C) The equation $6x^2+xy-y^2=0$ can be factored as $-(y-3x)(y+2x)=0$,which gives the lines $y=3x$ and $y=-2x$.To find the vertices of the triangle,we

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$\left(\frac{-1}{3}, \frac{7}{3}\right)$

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(C) The equation $6x^2+xy-y^2=0$ can be factored as $-(y-3x)(y+2x)=0$,which gives the lines $y=3x$ and $y=-2x$.To find the vertices of the triangle,we find the intersection points of these lines with $x+3y=10$:$1$. Intersection of $y=3x$ and $x+3y=10$: $x+3(3x)=10 \implies 10x=10 \implies x=1, y=3$. Vertex is $(1,3)$.$2$. Intersection of $y=-2x$ and $x+3y=10$: $x+3(-2x)=10 \implies -5x=10 \implies x=-2, y=4$. Vertex is $(-2,4)$.$3$. Intersection of $y=3x$ and $y=-2x$ is the origin $(0,0)$.The centroid is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{0+1-2}{3}, \frac{0+3+4}{3}\right) = \left(-\frac{1}{3}, \frac{7}{3}\right)$.

The centroid of the triangle formed by the lines $x+3y=10$ and $6x^2+xy-y^2=0$ is

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