If one of the lines given by 6x^2 - xy + 4cy^ | vedclass

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(A) The given equation of the pair of lines is $6x^2 - xy + 4cy^2 = 0$ $(i)$.Since $3x + 4y = 0$ is a line represented by this equation,its slope is $

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$-3$

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(A) The given equation of the pair of lines is $6x^2 - xy + 4cy^2 = 0$ $(i)$.Since $3x + 4y = 0$ is a line represented by this equation,its slope is $m_1 = -\frac{3}{4}$.For a homogeneous equation $ax^2 + 2hxy + by^2 = 0$,the product of the slopes of the lines is $m_1 m_2 = \frac{a}{b}$.Here,$a = 6$,$2h = -1$,and $b = 4c$. Thus,$m_1 m_2 = \frac{6}{4c} = \frac{3}{2c}$.Substituting $m_1 = -\frac{3}{4}$,we get $(-\frac{3}{4}) m_2 = \frac{3}{2c}$,which implies $m_2 = -\frac{2}{c}$.The sum of the slopes is $m_1 + m_2 = -\frac{2h}{b} = -\frac{-1}{4c} = \frac{1}{4c}$.Substituting the values of $m_1$ and $m_2$: $-\frac{3}{4} - \frac{2}{c} = \frac{1}{4c}$.Multiplying by $4c$,we get $-3c - 8 = 1$.$-3c = 9$,so $c = -3$.

If one of the lines given by $6x^2 - xy + 4cy^2 = 0$ is $3x + 4y = 0$,then $c$ equals

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