The slope of a line L is 2. If m_1 and m_2 ar | vedclass

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(D) The angle $	heta$ between two lines with slopes $m$ and $m'$ is given by $	an 	heta = \left| \frac{m - m'}{1 + m m'} ight|$.Given $	heta = \

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$-16$

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(D) The angle $	heta$ between two lines with slopes $m$ and $m'$ is given by $	an 	heta = \left| \frac{m - m'}{1 + m m'} ight|$.Given $	heta = \frac{\pi}{6}$ and $m' = 2$,we have $	an \frac{\pi}{6} = \left| \frac{m - 2}{1 + 2m} ight|$.Since $	an \frac{\pi}{6} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \left| \frac{m - 2}{1 + 2m} ight|$.Squaring both sides,we get $\frac{1}{3} = \frac{(m - 2)^2}{(1 + 2m)^2}$.$(1 + 2m)^2 = 3(m - 2)^2$.$1 + 4m + 4m^2 = 3(m^2 - 4m + 4)$.$1 + 4m + 4m^2 = 3m^2 - 12m + 12$.$m^2 + 16m - 11 = 0$.This is a quadratic equation in $m$ whose roots are $m_1$ and $m_2$.Using the sum of roots formula,$m_1 + m_2 = -\frac{b}{a} = -\frac{16}{1} = -16$.

The slope of a line $L$ is $2$. If $m_1$ and $m_2$ are the slopes of two lines which are inclined at an angle of $\frac{\pi}{6}$ with $L$,then $m_1 + m_2 =$

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