KCET · Maths · Straight Lines
The angle between the line \(x+y=3\) and the line joining the points \((1,1)\) and \((-3,4)\) is
- A \(\tan ^{-1}(7)\)
- B \(\tan ^{-1}\left(-\frac{1}{7}\right)\)
- C \(\tan ^{-1}\left(\frac{1}{7}\right)\)
- D \(\tan ^{-1}\left(\frac{2}{7}\right)\)
Answer & Solution
Correct Answer
(C) \(\tan ^{-1}\left(\frac{1}{7}\right)\)
Step-by-step Solution
Detailed explanation
\(\because y=-x+3 ; m_1=-1=\tan \theta_1\)
and \(m_2=\frac{y-1}{x-1}=\frac{4-1}{-3-1}\)
\(\frac{y-1}{x-1}=\frac{3}{-4} ; m_2=\frac{3}{-4}=\tan \theta_2\)
Angle between two lines are
\(=\tan \left|\theta_2-\theta_1\right|=\frac{\tan \theta_2-\tan \theta_1}{1+\tan \theta_1 \cdot \tan \theta_2}=\frac{m_2 \div m_1}{1+m_1 m_2}\)
\(=\frac{\frac{-3}{4}+1}{1+\frac{3}{4}}=\frac{1}{4} \times \frac{4}{7}\)
Hence, \(\left|\theta_2-\theta_1\right|=\tan ^{-1}\left(\frac{1}{7}\right)\)
and \(m_2=\frac{y-1}{x-1}=\frac{4-1}{-3-1}\)
\(\frac{y-1}{x-1}=\frac{3}{-4} ; m_2=\frac{3}{-4}=\tan \theta_2\)
Angle between two lines are
\(=\tan \left|\theta_2-\theta_1\right|=\frac{\tan \theta_2-\tan \theta_1}{1+\tan \theta_1 \cdot \tan \theta_2}=\frac{m_2 \div m_1}{1+m_1 m_2}\)
\(=\frac{\frac{-3}{4}+1}{1+\frac{3}{4}}=\frac{1}{4} \times \frac{4}{7}\)
Hence, \(\left|\theta_2-\theta_1\right|=\tan ^{-1}\left(\frac{1}{7}\right)\)
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