Question

Find the equation of the straight lines passing through the origin and making an angle of 45° with the straight line $\sqrt{3}x+y=11$.

Answer 1

We know that, the equations of two lines passing through a point $\left(x_1,y_1\right)$ and making an angle $\mathrm{\alpha }$ with the given line y = mx + c are

$y-y_1=\frac{m\pm \tan \alpha }{1\mp m\tan \alpha }\left(x-x_1\right)$

Here,
$x_1=0,y_1=0,\mathrm{\alpha }={45}^{\circ }\mathrm{and}m=-\sqrt{3}$

So, the equations of the required lines are

$$\begin{gathered} y-0=\frac{-\sqrt{3}+\tan {45}^{\circ }}{1+\sqrt{3}\tan {45}^{\circ }}\left(x-0\right)\mathrm{and}y-0=\frac{-\sqrt{3}-\tan {45}^{\circ }}{1-\sqrt{3}\tan {45}^{\circ }}\left(x-0\right) \\ \Rightarrow y=\frac{-\sqrt{3}+1}{1+\sqrt{3}}x\mathrm{and}y=\frac{\sqrt{3}+1}{\sqrt{3}-1}x \\ \Rightarrow y=-\frac{3+1-2\sqrt{3}}{3-1}x\mathrm{and}y=\frac{3+1+2\sqrt{3}}{3-1}x \\ \Rightarrow y=\left(\sqrt{3}-2\right)x\mathrm{and}y=\left(\sqrt{3}+2\right)x \end{gathered}$$