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Find the equations of the straight lines passing through the point \mathrm{(2,3)} and inclined at \mathrm{\pi / 4}  radians to the line \mathrm{2 x+3 y=5}.

 

Option: 1

\mathrm{x-4 y+13=0 \: and\: 4 x+y-13=0}
 


Option: 2

\mathrm{x-5 y+13=0 \: and \: 5 x+y-13=0}
 


Option: 3

\mathrm{\quad x-5 y-13=0 \: and \: 5 x-y+13=0}
 


Option: 4

\mathrm{x+y+12=0 \: and \: x-y-4=0}


Answers (1)

best_answer

Let the line \mathrm{2 x+3 y=5} makes angle \mathrm{\theta} with positive \mathrm{x}-axis. Then \mathrm{\tan \theta=-2 / 3.}

Now \mathrm{\tan \theta \cdot \tan (\pi / 4)=-\frac{2}{3} \times \tan \left(\frac{\pi}{4}\right)=-\frac{2}{3} \neq \pm 1}

Slopes of the required lines are

\mathrm{ \tan (\theta+\pi / 4)=\frac{\tan \theta+\tan \frac{\pi}{4}}{1-\tan \theta \tan \frac{\pi}{4}}=\frac{-\frac{2}{3}+1}{1-\left(-\frac{2}{3}\right)}=\frac{1}{5} }

and \mathrm{\tan (\theta-\pi / 4)=-5 }

\therefore the equations of the required lines are

\mathrm{ \begin{array}{lll} y-3 & =\frac{1}{5}(x-2) \\ y-3 & =-5(x-2) \end{array} \quad \text { i.e. } \quad \begin{aligned} & \\ y-5 y+13 & =0 \\ 5 x+y-13 & =0 \end{aligned} }

Hence option 2 is correct.


 

Posted by

Ajit Kumar Dubey

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