A) \[21x+77y-101=0\]
B) \[11x-3y+9=0\]
C) \[31x+77y+101=0\]
D) \[11x-3y-9=0\]
Correct Answer: B
Solution :
Bisectors of angles is given by \[\frac{3x-4y+7}{5}=\pm \frac{12x+5y-2}{13}\] Þ \[11x-3y+9=0\] ......(i) and \[21x+77y-101=0\] ......(ii) Let the angle between the line \[3x-4y+7=0\]and (i) is \[\alpha ,\] then \[\tan \alpha =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{\frac{3}{4}-\frac{11}{3}}{1+\frac{3}{4}\times \frac{11}{3}} \right|=\frac{35}{45}<1\] \[\Rightarrow \alpha <{{45}^{o}}\] Hence \[11x-3y+9=0\]is the bisector of the acute angle between the given lines.
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