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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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Let slope of line, which inclined with angle $60^{\circ}$ with line $L$ is ' $n$ ', so
$$
\begin{aligned}
& \tan 60^{\circ}=\left|\frac{n-1}{1+n}\right|=\sqrt{3} \\
& \Rightarrow \quad(n-1)^2=3(n+1)^2 \\
& \Rightarrow 2 n^2+8 n+2=0 \text {, which roots are slope of } \\
& \text { required lines, so product of slopes }=1 \text {. } \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \tan 60^{\circ}=\left|\frac{n-1}{1+n}\right|=\sqrt{3} \\
& \Rightarrow \quad(n-1)^2=3(n+1)^2 \\
& \Rightarrow 2 n^2+8 n+2=0 \text {, which roots are slope of } \\
& \text { required lines, so product of slopes }=1 \text {. } \\
&
\end{aligned}
$$
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