If m=1 is the slope of a line L,then the prod | vedclass

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(A) Let the slope of the lines inclined at an angle of $60^{\circ}$ with line $L$ be $n$. Given the slope of line $L$ is $m=1$. The angle $	heta$ bet

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(A) Let the slope of the lines inclined at an angle of $60^{\circ}$ with line $L$ be $n$. Given the slope of line $L$ is $m=1$. The angle $	heta$ between two lines with slopes $m$ and $n$ is given by $	an 	heta = \left| \frac{n-m}{1+nm} ight|$. Substituting the values,we have $	an 60^{\circ} = \left| \frac{n-1}{1+n} ight| = \sqrt{3}$. Squaring both sides,we get $\frac{(n-1)^2}{(n+1)^2} = 3$. $(n-1)^2 = 3(n+1)^2 $ $\Rightarrow n^2 - 2n + 1 = 3(n^2 + 2n + 1) $ $\Rightarrow n^2 - 2n + 1 = 3n^2 + 6n + 3 $ $\Rightarrow 2n^2 + 8n + 2 = 0 $ $\Rightarrow n^2 + 4n + 1 = 0$. The roots of this quadratic equation represent the slopes of the two lines. The product of the roots (slopes) is given by the constant term divided by the coefficient of $n^2$,which is $\frac{1}{1} = 1$.

If $m=1$ is the slope of a line $L$,then the product of the slopes of non-parallel lines which are inclined at an angle of $60^{\circ}$ with $L$ is

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