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(C) The given pair of lines is $2y^2+5xy-3x^2=0$,which can be rewritten as $3x^2-5xy-2y^2=0$.Factoring the quadratic equation: $3x^2-6xy+xy-2y^2=0$ $\

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(C) The given pair of lines is $2y^2+5xy-3x^2=0$,which can be rewritten as $3x^2-5xy-2y^2=0$.Factoring the quadratic equation: $3x^2-6xy+xy-2y^2=0$ $\Rightarrow 3x(x-2y)+y(x-2y)=0$ $\Rightarrow (x-2y)(3x+y)=0$.Thus,the two lines are $L_1: x-2y=0$ and $L_2: 3x+y=0$.The third line is $L_3: x+y=k$.The vertices of the triangle are the intersection points of these lines:$O(0,0)$ is the intersection of $L_1$ and $L_2$.Intersection of $L_1$ and $L_3$: $x-2y=0$ and $x+y=k$ $\Rightarrow 3y=k$ $\Rightarrow y=\frac{k}{3}, x=\frac{2k}{3}$. So,$A(\frac{2k}{3}, \frac{k}{3})$.Intersection of $L_2$ and $L_3$: $3x+y=0$ and $x+y=k$ $\Rightarrow 2x=-k$ $\Rightarrow x=-\frac{k}{2}, y=\frac{3k}{2}$. So,$B(-\frac{k}{2}, \frac{3k}{2})$.The centroid of $	riangle OAB$ is $(\frac{0+\frac{2k}{3}-\frac{k}{2}}{3}, \frac{0+\frac{k}{3}+\frac{3k}{2}}{3}) = (\frac{\frac{k}{6}}{3}, \frac{\frac{11k}{6}}{3}) = (\frac{k}{18}, \frac{11k}{18})$.Given the centroid is $(\frac{1}{18}, \frac{11}{18})$,we have $\frac{k}{18} = \frac{1}{18}$,which implies $k=1$.

If the centroid of the triangle formed by the lines $2y^2+5xy-3x^2=0$ and $x+y=k$ is $(\frac{1}{18}, \frac{11}{18})$,then the value of $k$ equals $..........$

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