Question

The angle between the lines $y=(2-\sqrt{3})x+5$ and $y=\left(2+\sqrt{3}\right)x-7$ is

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

Answer 1

The correct option is B

$60°$

Finding the angles between the lines:

Given, the line of equation as $y=(2-\sqrt{3})x+5$ and $y=\left(2+\sqrt{3}x\right)-7$

The equation can be written as

$(2-\sqrt{3})x-y+5=0\to \left(i\right)$

and

$\left(2+\sqrt{3}x\right)-y-7=0\to \left(ii\right)$

The slope of the general equation $ax+by+c=0$ as $m=\frac{-a}{b}$

Finding the slope of the equation $\left(i\right)$

$a=2-\sqrt{3}$ $b=-1$

$$\begin{gathered} m_1=\frac{-{\displaystyle \left(2-\sqrt{3}\right)}}{{\displaystyle -1}} \\ {m}_1=2-\sqrt{3} \end{gathered}$$

Finding the slope of the equation $\left(ii\right)$

$a=2+\sqrt{3}$, $b=-1$

$$\begin{gathered} m_2=\frac{-{\displaystyle \left(2+\sqrt{3}\right)}}{{\displaystyle -1}} \\ {m}_2=2+\sqrt{3} \end{gathered}$$

We know that ,

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

$\begin{array}{l}\tan \theta =\left|\frac{\left(2-\sqrt{3}\right)-\left(2+\sqrt{3}\right)}{1+\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\right|\\ =\left|\frac{2-\sqrt{3}-2-\sqrt{3}}{1+4-3}\right|\\ =\left|\frac{-2\sqrt{3}}{2}\right|\\ =\left|-\sqrt{3}\right|\\ =\sqrt{3}\end{array}$

Now, finding the value of $\theta$

$$\begin{gathered} \Rightarrow \tan \theta =\sqrt{3} \\ \theta ={\tan }^{-1}\left(\sqrt{3}\right) \\ \theta =60° \end{gathered}$$

Therefore, the angle between the two lines is $60°$.

Hence, option (B) is the correct answer.