The straight line passing through (-1, 1) and | vedclass

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Question

(B) Factorizing the first pair of lines: $6x^2 - xy - 12y^2 = 6x^2 - 9xy + 8xy - 12y^2 = 3x(2x - 3y) + 4y(2x - 3y) = (3x + 4y)(2x - 3y) = 0$ Factorizi

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$3x + 4y - 1 = 0$

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(B) Factorizing the first pair of lines: $6x^2 - xy - 12y^2 = 6x^2 - 9xy + 8xy - 12y^2 = 3x(2x - 3y) + 4y(2x - 3y) = (3x + 4y)(2x - 3y) = 0$ Factorizing the second pair of lines: $15x^2 + 14xy - 8y^2 = 15x^2 + 20xy - 6xy - 8y^2 = 5x(3x + 4y) - 2y(3x + 4y) = (5x - 2y)(3x + 4y) = 0$ The common line is $3x + 4y = 0$. The slope of this line is $m = -\frac{3}{4}$. The equation of the line passing through $(-1, 1)$ with slope $m = -\frac{3}{4}$ is: $y - 1 = -\frac{3}{4}(x + 1)$ $4y - 4 = -3x - 3$ $3x + 4y - 1 = 0$

The straight line passing through $(-1, 1)$ and parallel to the common line of the pairs of lines given by $6x^2 - xy - 12y^2 = 0$ and $15x^2 + 14xy - 8y^2 = 0$ is:

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