The centroid of the triangle formed by the li | vedclass

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(C) The given pair of lines is $6x^2 + xy - y^2 = 0$. Factoring the quadratic expression: $6x^2 + 3xy - 2xy - y^2 = 0$ $\Rightarrow 3x(2x + y) - y(2x

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

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(C) The given pair of lines is $6x^2 + xy - y^2 = 0$. Factoring the quadratic expression: $6x^2 + 3xy - 2xy - y^2 = 0$ $\Rightarrow 3x(2x + y) - y(2x + y) = 0$ $\Rightarrow (3x - y)(2x + y) = 0$. Thus,the two lines are $L_1: 3x - y = 0$ and $L_2: 2x + y = 0$. The third line is $L_3: x + 3y = 10$. Finding the vertices of the triangle: $1$. Intersection of $L_1$ and $L_2$: Solving $3x - y = 0$ and $2x + y = 0$ gives $(0, 0)$. $2$. Intersection of $L_2$ and $L_3$: Solving $2x + y = 0 \Rightarrow y = -2x$ and $x + 3(-2x) = 10$ $\Rightarrow -5x = 10$ $\Rightarrow x = -2, y = 4$. Vertex is $(-2, 4)$. $3$. Intersection of $L_3$ and $L_1$: Solving $3x - y = 0 \Rightarrow y = 3x$ and $x + 3(3x) = 10$ $\Rightarrow 10x = 10$ $\Rightarrow x = 1, y = 3$. Vertex is $(1, 3)$. The centroid $(G)$ is given by $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) = \left(\frac{0 - 2 + 1}{3}, \frac{0 + 4 + 3}{3}\right) = \left(-\frac{1}{3}, \frac{7}{3}\right)$.

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

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