Question

Find the angle between the lines $y-\sqrt{3}x-5=0$ and $\sqrt{3}y-x+6=0$.

• A. $30^{\circ }$
• B. $150^{\circ }$
• C. $120^{\circ }$
• D. Either $(a)$ or $(b)$

Answer 1

Answer: (D)

Given lines are
$y-\sqrt{3}x-5=0$ or $y=\sqrt{3}x+5\dots \left(i\right)$
and $\sqrt{3}y-x+6=0$ or $y=\frac{1}{\sqrt{3}}x-2\sqrt{3}\dots \left(ii\right)$
Slope of line $\left(i\right)$ is $m_1=\sqrt{3}$ and slope of line $\left(ii\right)$ is $m_2=\frac{1}{\sqrt{3}}$.
The acute angle (say $\theta$)between two lines is given by
$tan\theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|=\left|\frac{\frac{1}{\sqrt{3}}-\sqrt{3}}{1+\sqrt{3}\times \frac{1}{\sqrt{3}}}\right|$
$=\left|\frac{1-3}{2\sqrt{3}}\right|=\pm \frac{1}{\sqrt{3}}$
which gives $\theta =30^{\circ }$. Hence, the angle between two lines is either $30^{\circ }$ or ${\left(180-30\right)}^{\circ }=150^{\circ }$