If m_1 and m_2 (m_1 m_2) are the slopes of | vedclass

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(C) The slope of the line joining the points $(1, 2)$ and $(3, 4)$ is $m = \frac{4-2}{3-1} = \frac{2}{2} = 1$. Let the slopes of the required lines be

Vedclass · Accepted Answer

$7+4\sqrt{3}$

Vedclass · Answer

(C) The slope of the line joining the points $(1, 2)$ and $(3, 4)$ is $m = \frac{4-2}{3-1} = \frac{2}{2} = 1$. Let the slopes of the required lines be $m_1$ and $m_2$. The angle between these lines and the given line is $30^{\circ}$. Using the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$,we have $	an 30^{\circ} = \left| \frac{m - 1}{1 + m(1)} ight| = \left| \frac{m - 1}{1 + m} ight|$. $\frac{1}{\sqrt{3}} = \left| \frac{m - 1}{m + 1} ight| \Rightarrow \pm \frac{1}{\sqrt{3}} = \frac{m - 1}{m + 1}$. Case $1$: $\frac{1}{\sqrt{3}} = \frac{m - 1}{m + 1}$ $\Rightarrow m + 1 = \sqrt{3}m - \sqrt{3}$ $\Rightarrow m(\sqrt{3} - 1) = \sqrt{3} + 1$ $\Rightarrow m = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 1 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$. Case $2$: $-\frac{1}{\sqrt{3}} = \frac{m - 1}{m + 1}$ $\Rightarrow -m - 1 = \sqrt{3}m - \sqrt{3}$ $\Rightarrow m(\sqrt{3} + 1) = \sqrt{3} - 1$ $\Rightarrow m = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{3 + 1 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$. Since $m_1 > m_2$,we have $m_1 = 2 + \sqrt{3}$ and $m_2 = 2 - \sqrt{3}$. Therefore,$\frac{m_1}{m_2} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} = \frac{(2 + \sqrt{3})^2}{4 - 3} = 4 + 3 + 4\sqrt{3} = 7 + 4\sqrt{3}$.

If $m_1$ and $m_2$ $(m_1 > m_2)$ 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} = $

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