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 \[2x-4y=0\]
Correct Answer: A
Solution :
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 \[x+3y=0\]
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