Question

The slope of a line is double of the slope of another line. If tangent of the angle between them is $\frac{1}{3}$, find the slopes of the lines.

Answer 1

Let the slopes of two lines be m and 2m.

It is given that $tan\theta =\frac{1}{3}$

$\therefore \frac{1}{3}=\left|\frac{m-2m}{1+m\times 2m}\right|\Rightarrow \frac{1}{3}=\left|\frac{-m}{1+2m^2}\right|$

$\therefore$ Slope of AB = $tan\theta$

Now $\frac{-m}{1+2m^2}=\frac{1}{3}\Rightarrow -3m=1+2m^2$

$\Rightarrow 2m^2+3m+1=0\Rightarrow -(m+1)(2m+1)=0$

$\Rightarrow m=-1$ and $\frac{-1}{2}$

when $\frac{-m}{1+2m^2}=\frac{-1}{3}\Rightarrow -3m=-1-2m^2$

$\Rightarrow 2m^2-3m+1=0$

$\Rightarrow (m-1)(2m-1)=0\Rightarrow m=1$ and $\frac{1}{2}$

Thus the slopes of lines are -1 and $-\frac{1}{2}$ or 1 and $\frac{1}{2}$