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

$\begin{array}{1 1}(A)\;-3\\(B)\;-1\\(C)\;3\\(D)\;1 \end{array}$

$3x+4y=0$
is one of the lines of the pair.
$6x^2-xy+4cy^2=0$
Put $y=-\large\frac{-3}{4}$$x We get 6x^2+\large\frac{3}{4}$$x^2+4c\bigg(\large\frac{-3}{4}$$x^2 \bigg)=0 => 6+\large\frac{3}{4} +\frac{9c}{4}$$=0$
$c=-3$
Hence A is the correct answer.