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Question
Find the angles between the following pair of straight lines:
3x − y + 5 = 0 and x − 3y + 1 = 0
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Solution
The equations of the lines are
3x − y + 5 = 0 ... (1)
x − 3y + 1 = 0 ... (2)
Let \[m_1 \text { and } m_2\] be the slopes of these lines.
\[m_1 = 3, m_2 = \frac{1}{3}\]
Let \[\theta\] be the angle between the lines.
Then,
\[\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|\]
\[ = \left| \frac{3 - \frac{1}{3}}{1 + 1} \right|\]
\[ = \frac{4}{3}\]
\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{4}{3} \right)\]
Hence, the acute angle between the lines is \[\tan^{- 1} \left( \frac{4}{3} \right)\].
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