If one of the lines given by $6 x^2-x y+4 c y^2=0$ is $3 x+4 y=0$, then $c$ equals
If one of the lines given by $6 x^2-x y+4 c y^2=0$ is $3 x+4 y=0$, then $c$ equals
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1
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$-1$
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3
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$-3$
Solution
$m_1+m_2=\frac{1}{4 c}, m_1 m_2=\frac{6}{4 c}$ and $m_1=-\frac{3}{4}$ Hence c $=-3$
Asked in: JEE Main 2004
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