If L is a line passing through the point (-1, | vedclass

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(C) First,we factorize the given pairs of lines to find the common line.\$6x^2 - xy - 12y^2 = 6x^2 - 9xy + 8xy - 12y^2 = 3x(2x - 3y) + 4y(2x - 3y) = (

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$x^2 - y^2 = 0$

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(C) First,we factorize the given pairs of lines to find the common line.\$6x^2 - xy - 12y^2 = 6x^2 - 9xy + 8xy - 12y^2 = 3x(2x - 3y) + 4y(2x - 3y) = (3x + 4y)(2x - 3y) = 0$.\$15x^2 + 14xy - 8y^2 = 15x^2 + 20xy - 6xy - 8y^2 = 5x(3x + 4y) - 2y(3x + 4y) = (5x - 2y)(3x + 4y) = 0$.\The common line is $3x + 4y = 0$.\Line $L$ passes through $(-1, 1)$ and is parallel to $3x + 4y = 0$,so its equation is $3x + 4y + k = 0$. Substituting $(-1, 1)$,we get $3(-1) + 4(1) + k = 0$ $\Rightarrow -3 + 4 + k = 0$ $\Rightarrow k = -1$. Thus,$L: 3x + 4y - 1 = 0$,or $3x + 4y = 1$.\To find the pair of lines joining the origin to the intersection of $2x^2 - xy - y^2 + x - y = 0$ and $3x + 4y = 1$,we homogenize the curve equation using $1 = 3x + 4y$:\$2x^2 - xy - y^2 + (x - y)(3x + 4y) = 0$.\Expanding this: $2x^2 - xy - y^2 + 3x^2 + 4xy - 3xy - 4y^2 = 0$.\Combining like terms: $(2 + 3)x^2 + (-1 + 4 - 3)xy + (-1 - 4)y^2 = 0$.\$5x^2 - 5y^2 = 0 \Rightarrow x^2 - y^2 = 0$.

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