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.