Question

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

• A. $90°$
• B. $60°$
• C. $45°$
• D. $30°$

Answer 1

The correct option is D

$30°$

Explanation for the correct option:

Step 1: Finding the value of $m_1$ and $m_2$

Let $Q_1=\sqrt{3}x-y-2$ be the first line and $Q_2=x-\sqrt{3}y+1$be the second line.

Now to find the slope of the line $Q_1$

$m_1=\frac{-a}{b}$

Substitute $a=\sqrt{3}$ and $b=-1$ in $m_1$

$$\begin{gathered} m_1=\frac{-\sqrt{3}}{-1} \\ {m}_1=\sqrt{3} \end{gathered}$$

Also to find the slope of the line $Q_2$

$m_2=\frac{-a}{b}$

Substitute $a=1$ and $b=-\sqrt{3}$ in $m_2$

$$\begin{gathered} m_2=\frac{-1}{-\sqrt{3}} \\ {m}_1=\frac{1}{\sqrt{3}} \end{gathered}$$

Step 2: Finding the angle

Let $\theta$ be the angle between the two lines and we know that:

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

Substitute $m_1=\sqrt{3}$ and $m_2=\frac{1}{\sqrt{3}}$

$$\begin{gathered} \tan \theta =\left|\frac{\sqrt{3}-{\displaystyle \frac{1}{\sqrt{3}}}}{1+\left(\sqrt{3}\right)\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}\right| \\ =\left|\frac{{\displaystyle \frac{{\left(\sqrt{3}\right)}^2-1}{\sqrt{3}}}}{1+1}\right| \\ =\left|\frac{3-1}{2\sqrt{3}}\right| \\ =\left|\frac{2}{2\sqrt{3}}\right| \\ =\left|\frac{1}{\sqrt{3}}\right| \end{gathered}$$

Let us consider the positive value of mod

$$\begin{gathered} \tan \theta =\frac{1}{\sqrt{3}} \\ \theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ \theta =30° \end{gathered}$$

Hence, the angle between the two lines is $30°$

Therefore, option (D) is the correct answer.