1.

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



Discussion

No Comment Found

Related InterviewSolutions