If the equation of the pair of straight lines | vedclass

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(C) The given pair of lines is $3x^2 + 11xy - 4y^2 = 0$. Factoring this,we get $(3x - y)(x + 4y) = 0$. The slopes of these lines are $m_1 = 3$ and $m_

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$-19$

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(C) The given pair of lines is $3x^2 + 11xy - 4y^2 = 0$. Factoring this,we get $(3x - y)(x + 4y) = 0$. The slopes of these lines are $m_1 = 3$ and $m_2 = -\frac{1}{4}$. The lines perpendicular to these will have slopes $m_1' = -\frac{1}{3}$ and $m_2' = 4$. Since these lines pass through $(1, 1)$,their equations are: $y - 1 = -\frac{1}{3}(x - 1) \Rightarrow x + 3y - 4 = 0$ $y - 1 = 4(x - 1) \Rightarrow 4x - y - 3 = 0$ The combined equation is $(x + 3y - 4)(4x - y - 3) = 0$. Expanding this: $4x^2 - xy - 3x + 12xy - 3y^2 - 9y - 16x + 4y + 12 = 0$. $4x^2 + 11xy - 3y^2 - 19x - 5y + 12 = 0$. Comparing with $ax^2 + 2hxy + by^2 + 2gx + 2fy + 12 = 0$,we get $a = 4, 2h = 11, b = -3, 2g = -19, 2f = -5$. Thus,$a = 4, h = \frac{11}{2}, b = -3, g = -\frac{19}{2}, f = -\frac{5}{2}$. Calculating $2(a - h + b - g + f - 12) = 2(4 - \frac{11}{2} - 3 + \frac{19}{2} - \frac{5}{2} - 12) = 2(1 - \frac{11}{2} + \frac{19}{2} - \frac{5}{2} - 12) = 2(1 + \frac{3}{2} - 12) = 2(\frac{5}{2} - 12) = 5 - 24 = -19$.

If the equation of the pair of straight lines passing through the point $(1, 1)$ and perpendicular to the pair of lines $3x^2 + 11xy - 4y^2 = 0$ is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ (where the constant term is $12$),then find the value of $2(a - h + b - g + f - 12)$.

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