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=

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$2 - \sqrt{3}$

$7 - 4 \sqrt{3}$

$2 + \sqrt{3}$

$7 + 4 \sqrt{3}$

answer is C.

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Detailed Solution

$\begin{array}{l} L e t θ = 30^{0} \\ L e t A (1, 2) B (3, 4) \\ m_{1} = S l o p e o f A B = \frac{4 - 2}{3 - 2} = 1 \\ L e t m_{2} = m i s t h e s l o p e o f a l i n e w h i c h m a k e s 30^{0} \\ t o t h e l i n e j o i n i n g (1, 2); (3, 4) \\ ∴ Tan 30^{0} = |\frac{1 - m}{1 + 1 (m)}| \\ \Rightarrow \frac{1}{\sqrt{3}} = |\frac{1 - m}{1 + m}| \\ \Rightarrow {(m + 1)}^{2} = 3 {(1 - m)}^{2} \\ \Rightarrow m^{2} + 1 + 2 m = 0 \Rightarrow m^{2} - 4 m + 1 = 0 \end{array}$

$\begin{array}{l} ∴ m = \frac{4 \pm \sqrt{16 - 4}}{2 (1)} = \frac{4 \pm 2 \sqrt{3}}{2} \Rightarrow 2 \pm \sqrt{3} \\ m_{1} = 2 + \sqrt{3} a n d m_{2} = 2 - \sqrt{3} \\ \frac{m_{1}}{m_{2}} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} \\ \Rightarrow \frac{{(2 + \sqrt{3})}^{2}}{4 - 3} \Rightarrow \frac{4 + 3 + 4 \sqrt{3}}{1} \Rightarrow 7 + 4 \sqrt{3} \end{array}$