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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.
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Solution
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 \[\pm \frac{1}{2}, \pm 1\] .
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