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  • Find the bisector of acute angle between the lines and <img alt="\mathrm{5 x+12

Find the bisector of acute angle between the lines \mathrm{3 x-4 y=0} and \mathrm{5 x+12 y+7=0},
 

Option: 1

\mathrm{2 x-16 y-5=0}


Option: 2

\mathrm{64 x+8 y+35=0}


Option: 3

\mathrm{2 x+8 y+35=0}


Option: 4

\mathrm{64 x-16 y-5=0}


Answers (1)

best_answer

Equations of the bisectors are

\mathrm{\frac{3 x-4 y}{5}= \pm \frac{5 x+12 y+7}{13} }
i.e. \mathrm{2 x-16 y-5=0}  and \mathrm{64 x+8 y+35=0}

Now suppose \theta be the angle between the given lines which is bisected by the bisector \mathrm{2 x-16 y-5=0}.

The angle between \mathrm{3 x-4 y=0} and \mathrm{2 x-16 y-5=0} is \mathrm{\frac{\theta }{ 2}}  which is certainly acute.

\begin{array}{ll} \therefore \quad & \mathrm{\left|\tan \frac{\theta}{2}\right|=\left|\frac{\frac{3}{4}-\frac{1}{8}}{1+\frac{3}{4} \times \frac{1}{8}}\right|=\left|\frac{24-4}{32+3}\right|=\frac{20}{35}=\frac{4}{7}<1 }\\ \therefore \quad&\mathrm{ \frac{\theta}{2}<\frac{\pi}{4} \text { and so } \theta<\frac{\pi}{2}} \end{array}

Hence \mathrm{2 x-16 y-5=0} is the required bisector.

Posted by

shivangi.shekhar

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