Question

The angle between the lines $\sqrt{3}x-y-2=0$ and $x-\sqrt{3}y+1=0$ is:

• A. $90{}^{\circ }$
• B. ${60}^{\text{o}}$
• C. $45{}^{\circ }$
• D. $15{}^{\circ }$
• E. $30{}^{\circ }$

Answer 1

Answer: (E)

Let $m_1=$ slope of $\sqrt{3}x-y-2=0$ i.e., $m_1=\sqrt{3}$ and $m_2=$ slope of $x-\sqrt{3}y+1=0$ i.e., $m_2=\frac{1}{\sqrt{3}}$ $\therefore$ $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+1}\right|=\left|\frac{3-1}{2.\sqrt{3}}\right|$ $=\frac{1}{\sqrt{3}}$ $\therefore$ $\theta =30{}^{\circ }$