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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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If one of the lines given by $6x^2-xy+4cy^2=0$ is $3x+4y=0$ then c equals

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

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is one of the lines of the pair.
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$
Hence A is the correct answer.
answered Apr 8, 2014 by meena.p

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