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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

The set of lines $ax+by+c=0$ where $3a+2b+4c=0$ is concurrent at the point.

$\begin{array}{1 1}(A)\;(\frac{1}{2},\frac{2}{3}) \\(B)\;(\frac{3}{4},\frac{1}{2})\\(C)\;(\frac{2}{3},2)\\(D)\;(\frac{2}{5},\frac{1}{5}) \end{array}$

1 Answer

Given that,
$3a+2b+4c=0$
$\large\frac{3}{4}$$a+\large\frac{2b}{4}+\large\frac{4c}{c}$$=0$
=> $\large\frac{3}{4}$$a+\large\frac{1}{2}$$b+c=0$
=> The set of lines $ax+by+c=0$ passes through $(\large\frac{3}{4},\frac{1}{2})$
Hence B is the correct answer.
answered Apr 8, 2014 by meena.p
 

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