Question

If $m_{1}, m_{2}\left(m_{1}>m_{2}\right)$ are the slopes of the lines which make an angle of $30^{\circ}$ with the line joining the points $(1,2)$ and $(3,4)$, then $\frac{m_{1}}{m_{2}}=$

• A. $2+\sqrt{3}$
• B. $2-\sqrt{3}$
• C. $7+4 \sqrt{3}$
• D. $7-4 \sqrt{3}$

Answer 1

Answer: (C)

Slope of $B C=\frac{4-2}{3-1}=1$
Let the slope of $A B=m_{1}$
$\therefore \tan 30^{\circ}=\left|\frac{m_{1}-1}{1+m_{1}}\right| \Rightarrow \pm \frac{1}{\sqrt{3}}=\frac{m_{1}-1}{1+m_{1}}$
$m_{1}+1=\pm \sqrt{3}\left(m_{1}-1\right)$
$m_{1}+1=\sqrt{3}\left(m_{1}-1\right)$
$m_{1}(\sqrt{3}-1)=\sqrt{3}+1$
$m_{1}=\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}$
Similarly, we get $m_{2}=2-\sqrt{3}$
$\therefore \frac{m_{1}}{m_{2}}=\frac{2+\sqrt{3}}{2-\sqrt{3}}=7+4 \sqrt{3}$