Question

Find the equations to the straight lines which pass through the origin and are inclined at an angle of 75° to the straight line $x+y+\sqrt{3}\left(y-x\right)=a$.

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,
$$\begin{gathered} \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{line}\mathrm{is}, \\ x+y+\sqrt{3}\left(y-x\right)=a \\ \Rightarrow \left(\sqrt{3}+1\right)y=\left(\sqrt{3}-1\right)x+a \\ \Rightarrow y=\frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)}x+\frac{a}{\left(\sqrt{3}+1\right)} \\ \mathrm{Comparing}\mathrm{this}\mathrm{equation}\mathrm{with}\mathit{}y=mx+c \\ \mathrm{we}\mathrm{get}, \\ m=\frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)} \end{gathered}$$
$\therefore x_1=0,y_1=0,\mathrm{\alpha }={75}^{\circ },m=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$
and $\tan {75}^{\circ }=2+\sqrt{3}$

So, the equations of the required lines are

$$\begin{gathered} y-0=\frac{2-\sqrt{3}+\tan {75}^{\circ }}{1-\left(2-\sqrt{3}\right)\tan {75}^{\circ }}\left(x-0\right)\mathrm{and}y-0=\frac{2-\sqrt{3}-\tan {75}^{\circ }}{1+\left(2-\sqrt{3}\right)\tan {75}^{\circ }}\left(x-0\right) \\ \Rightarrow y=\frac{2-\sqrt{3}+2+\sqrt{3}}{1-\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}x\mathrm{and}y=\frac{2-\sqrt{3}-2-\sqrt{3}}{1+\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}x \\ \Rightarrow y=\frac{4}{1-1}x\mathrm{and}y=-\sqrt{3}x \\ \Rightarrow x=0\mathrm{and}\sqrt{3}x+y=0 \end{gathered}$$