

InterviewSolution
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The angle between the lines \(\sqrt{3}x +y=1\) and \(x+\sqrt{3}y=1\) is1. 30o2. 45o3. 60o4. none of these |
Answer» Correct Answer - Option 1 : 30o Concept: If two lines with slopes m1 and m2 make an angle θ with each other, then: ⇒ tan θ = \(\rm \left|\frac{m_1-m_2}{1+m_1m_2}\right|\). Calculation: The given lines are \(\sqrt{3}x +y=1\) and \(x+\sqrt{3}y=1\) Let m1 and m2 be the slopes of these lines, then \(m_1=-\frac{\sqrt{3}}{1}\)= \(-\sqrt{3}\) and \(m_2= -\frac{1}{\sqrt{3}}\) If θ is the acute angle between the lines the, ⇒ tan θ = |\(\frac{m_1-m_2}{1+m_1m_2}\)| = |\(\frac{-\sqrt{3}-(-\frac{1}{\sqrt{3}})}{1+(-\sqrt{3})(-\frac{1}{\sqrt{3}})}\)| = |\(\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+1}\)| = |\(\frac{-3+1}{2\sqrt{3}}\)| = |\(\frac{-2}{2\sqrt{3}}\)| = \(\frac{1}{\sqrt{3}}\) ⇒ tan θ = 30o |
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