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  • Find the equations of the straight lines passing through the point (2,3) and inclined at <img alt="\frac{\pi }{ 4}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7B%5Cpi%20%7D%7B%204

Find the equations of the straight lines passing through the point (2,3) and inclined at \frac{\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 \theta with

positive x-axis. Then \mathrm{\tan \theta=\frac{-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}}
\therefore the equations of the required lines are

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

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