• # question_answer Two straight lines ${{L}_{1}}=0$ and ${{L}_{2}}=0$ pass through the origin forming an angle of ${{\tan }^{-1}}\left( \frac{7}{9} \right)$with each other. If the ratio of the slopes of ${{L}_{2}}=0$ and ${{L}_{1}}=0$ is $\frac{9}{2}$,then their equations are A)  $3x+2y=0$ and $x+3y=0$ B)  $x=3y$ and $3x+2y=0$ C)  $2x-3y=0$ and $x+3y=0$ D)  $y-3x=0$ and $2x-4y=0$

Let the slope of ${{L}_{1}}=0$ be m, then the slope of ${{L}_{2}}=0$ be $\frac{9m}{2}$.                (by condition) $\because$     $\tan \,\theta \,\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$ Hence,             $\frac{7}{9}=\,\left| \frac{4c-\frac{9m}{2}}{1+4c\cdot \,\frac{9m}{2}} \right|=\,\left| \,\frac{-7m}{2+9{{m}^{2}}} \right|$ $\left[ \because \,\,\theta \,\,={{\tan }^{-1}}\left( \frac{7}{9} \right) \right]$ $\Rightarrow$ $-\frac{7m}{2+9{{m}^{2}}}=\pm \,\frac{7}{9}$ $\Rightarrow$            $m=-\frac{2}{3},\,-\frac{1}{3},\,\frac{2}{3}\,,\,\frac{1}{3}$ Hence, the euqation of lines are $3y=x$ and $2y=3x,$ $3y=2x$ and $y=3x$ $x+3y=0$ and $3x+2y=0$ $2x+3y=0$ and $x+3y=0$