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(C) The given pair of lines is $2y^2-xy-6x^2=0$.Factoring the equation:$2y^2-4xy+3xy-6x^2=0$$2y(y-2x)+3x(y-2x)=0$$(y-2x)(2y+3x)=0$Thus,the sides of th

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$\left(\frac{-5}{9}, \frac{11}{9}\right)$

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(C) The given pair of lines is $2y^2-xy-6x^2=0$.Factoring the equation:$2y^2-4xy+3xy-6x^2=0$$2y(y-2x)+3x(y-2x)=0$$(y-2x)(2y+3x)=0$Thus,the sides of the triangle are $x+y=1$,$y-2x=0$,and $2y+3x=0$.Solving these equations in pairs to find the vertices:$1$. $x+y=1$ and $y-2x=0$: Substituting $y=2x$ into $x+y=1$ gives $3x=1$,so $x=1/3$ and $y=2/3$. Vertex: $(1/3, 2/3)$.$2$. $x+y=1$ and $2y+3x=0$: Substituting $y=1-x$ into $2y+3x=0$ gives $2(1-x)+3x=0$,so $2+x=0$,$x=-2$ and $y=3$. Vertex: $(-2, 3)$.$3$. $y-2x=0$ and $2y+3x=0$: Clearly,the origin $(0,0)$ is the intersection. Vertex: $(0,0)$.The centroid $(G)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.$G = \left(\frac{1/3+0-2}{3}, \frac{2/3+0+3}{3}\right) = \left(\frac{-5/3}{3}, \frac{11/3}{3}\right) = \left(\frac{-5}{9}, \frac{11}{9}\right)$.

The centroid of the triangle formed by the lines $x+y=1$ and $2y^2-xy-6x^2=0$ is

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