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 he lines.

Answer 1

Let m1 and m be the slopes of the two given lines such that .m1=2m.

We know that if θ is the angle between the lines l1and l2 with slopes m1and m2, then $tanθ=|\frac{m_2-m_1}{1+m_1m_2}|$ .

It is given that the tangent of the angle between the two lines is $\frac{1}{3}$.

∴ $\frac{1}{3}$= $|\frac{m-2m}{1+(2m).m}|$

⇒ $\frac{1}{3}$= $|\frac{-m}{1+2m^2}|$

⇒ $\frac{1}{3}$= $\frac{-m}{1+2m^2}$ or $\frac{1}{3}$=- $(\frac{-m}{1+2m^2})$= $\frac{m}{1+2m^2}$

Case I

⇒ $\frac{1}{3}$= $(\frac{-m}{1+2m^2})$

$⇒ 1+2m^2=-3m$

$⇒2m^2+3m+1=0$

$⇒2m^2+2m+m+1=0$

$⇒ 2m(m+1)+1(m+1)=0$

$⇒ (m+1)(2m+1)=0$

$⇒ m=-1 \,or \,\,m=-\frac{1}{2}$

If m = -1, then the slopes of the lines are -1 and -2.

If m = $-\frac{1}{2}$ then the slopes of the lines are $-\frac{1}{2}$ and -1.

Case II

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

$⇒ 2m^2+1=3m$

$⇒2m^2-3m-m+1=0$

$⇒ 2m^2-2m-m+1=0$

$⇒ 2m(m-1)-1(m-1)=0$

$⇒ (m-1)(2m-1)=0$

$⇒ m=1 \,or \,\,m=\frac{1}{2}$

If m = 1, then the slopes of the lines are 1 and 2

If m = $\frac{1}{2}$, then the slopes of the lines are $\frac{1}{2}$ and 1

Hence, the slopes of the lines are -1 and -2 or $-\frac{1}{2}$ and -1 or 1 and 2 or $\frac{1}{2}$ and 1.