If m_1, m₂ (m₁ > m₂) are the slopes of the lines which make angle of 30⁰ with the line joining the points (1, 2) and (3, 4) , then m₁/m₂  =

$\begin{array}{l}Let\theta ={30}^0\\ LetA\left(1,2\right)B\left(3,4\right)\\ m_1=SlopeofAB=\frac{4-2}{3-2}=1\\ Let{m}_2=mistheslopeofalinewhichmakes{30}^0\\ tothelinejoining\left(1,2\right);\left(3,4\right)\\ \therefore \mathrm{Tan}{30}^0=\left|\frac{1-m}{1+1\left(m\right)}\right|\\ \Rightarrow \frac{1}{\sqrt{3}}=\left|\frac{1-m}{1+m}\right|\\ \Rightarrow {\left(m+1\right)}^2=3{\left(1-m\right)}^2\\ \Rightarrow m^2+1+2m=0\Rightarrow m^2-4m+1=0\end{array}$

$\begin{array}{l}\therefore m=\frac{4\pm \sqrt{16-4}}{2\left(1\right)}=\frac{4\pm 2\sqrt{3}}{2}\Rightarrow 2\pm \sqrt{3}\\ m_1=2+\sqrt{3}and{m}_2=2-\sqrt{3}\\ \frac{m_1}{m_2}=\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}\\ \Rightarrow \frac{{\left(2+\sqrt{3}\right)}^2}{4-3}\Rightarrow \frac{4+3+4\sqrt{3}}{1}\Rightarrow 7+4\sqrt{3}\end{array}$