Question

The slope of a line is double of the slope of another line. If tangents of the angle between them is \[\frac{1}{3}\],find the slopes of the other line.

Answer 1

Let \[m_1 \text { and  } m_2\] be the slopes of the given lines. 

\[\therefore m_2 = 2 m_1\]

Let \[\theta\] be the angle between the given lines.

\[\therefore \tan\theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|\]

\[ \Rightarrow \frac{1}{3} = \left| \frac{2 m_1 - m_1}{1 + 2 {m_1}^2} \right| = \left| \frac{m_1}{1 + 2 {m_1}^2} \right|\]

\[ \Rightarrow \frac{m_1}{1 + 2 {m_1}^2} = \pm \frac{1}{3}\]

Taking the positive sign, we get,

\[3 m_1 = 1 + 2 {m_1}^2 \]

\[ \Rightarrow 2 {m_1}^2 - 3 m_1 + 1 = 0\]

\[ \Rightarrow \left( 2 m_1 - 1 \right)\left( m_1 - 1 \right) = 0\]

\[ \Rightarrow m_1 = \frac{1}{2}, 1\]

Taking the negative sign, we get,

\[- 3 m_1 = 1 + 2 {m_1}^2 \]

\[ \Rightarrow 2 {m_1}^2 + 3 m_1 + 1 = 0\]

\[ \Rightarrow \left( 2 m_1 + 1 \right)\left( m_1 + 1 \right) = 0\]

\[ \Rightarrow m_1 = - \frac{1}{2}, - 1\]

Hence, the slopes of the other line are \[\pm \frac{1}{2}, \pm 1\] .