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  • Find the bisector of acute angle between the lines <img alt="\mathrm{3 x-4 y=0 \: and \: 5 x+12 y+7=0,}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B3%20x-4%20y%3D0%20%5C%3A%20

Find the bisector of acute angle between the lines \mathrm{3 x-4 y=0 \: and \: 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\: 64 x+8 y+35=0 }

Now suppose \mathrm{ \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\: 2 x-16 y-5=0} is \mathrm{\theta / 2} which is certainly acute.

\mathrm{\therefore \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 }

\mathrm{ \therefore \quad \frac{\theta}{2}<\frac{\pi}{4} \text { and so } \theta<\frac{\pi}{2} }

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

Hence option 1 is correct.

Posted by

vishal kumar

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