If the centroid of the triangle formed by the | vedclass

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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: $3x^2 - 6xy + xy -

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$1$

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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: $3x^2 - 6xy + xy - 2y^2 = 0$ $\Rightarrow 3x(x - 2y) + y(x - 2y) = 0$ $\Rightarrow (x - 2y)(3x + y) = 0$. The lines are $L_1: x - 2y = 0$ and $L_2: 3x + y = 0$. These lines intersect at the origin $O(0, 0)$. The third line is $L_3: x + y = k$. Intersection of $L_1$ and $L_3$: $x - 2y = 0$ and $x + y = k$. Solving gives $3y = k \Rightarrow y = \frac{k}{3}$ and $x = \frac{2k}{3}$. So,$A = \left(\frac{2k}{3}, \frac{k}{3}ight)$. Intersection of $L_2$ and $L_3$: $3x + y = 0$ and $x + y = k$. Subtracting gives $2x = -k \Rightarrow x = -\frac{k}{2}$ and $y = \frac{3k}{2}$. So,$B = \left(-\frac{k}{2}, \frac{3k}{2}ight)$. The centroid $G$ of $	riangle OAB$ is $\left(\frac{0 + \frac{2k}{3} - \frac{k}{2}}{3}, \frac{0 + \frac{k}{3} + \frac{3k}{2}}{3}ight) = \left(\frac{\frac{k}{6}}{3}, \frac{\frac{11k}{6}}{3}ight) = \left(\frac{k}{18}, \frac{11k}{18}ight)$. Given $G = \left(\frac{1}{18}, \frac{11}{18}ight)$,we equate the coordinates: $\frac{k}{18} = \frac{1}{18} \Rightarrow k = 1$.

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

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