Question

If the lines $y=(2+\sqrt{3})x+4$ and $y=kx+6$ are inclined at an angle $60°$ to each other, then the value of $k$ will be

• A. $1$
• B. $2$
• C. $-1$
• D. $-2$

Answer 1

The correct option is C

$-1$

Explanation for the correct option.

Step 1: Find the slope of the given lines.

In the question, the lines $y=(2+\sqrt{3})x+4$ and $y=kx+6$ are given.

We know that the slope-intercept form of a line is $y=mx+c$. where $(x,y)$ is the general point on the line, $m$ is the slope of the line and $c$ is the $y-$intercept.

So, the slope $m_1$ of the line $y=(2+\sqrt{3})x+4$ is $2+\sqrt{3}$.

And the slope $m_2$ of the line $y=kx+6$ is $k$.

Step 2: Find the value of $k$.

We know that, if $m_1$ and $m_2$ are the slopes of two line and $\theta$ is the angle between them, then $\tan \theta =\left|\frac{m_2-m_1}{1+m_1·m_2}\right|$.

Since the angle between the given lines is $60°$ and the slopes of the lines are $2+\sqrt{3}$ and $k$.

So,

$$\begin{gathered} \tan (60°)=\left|\frac{2+\sqrt{3}-k}{1+(2+\sqrt{3})k}\right| \\ \Rightarrow \sqrt{3}=\left|\frac{2+\sqrt{3}-k}{1+(2+\sqrt{3})k}\right| \\ \Rightarrow \sqrt{3}\left(1+(2+\sqrt{3})k\right)=\pm \left(2+\sqrt{3}-k\right) \\ \Rightarrow \sqrt{3}+\sqrt{3}(2+\sqrt{3})k=\pm \left(2+\sqrt{3}-k\right) \end{gathered}$$

When the left-hand side is equal to $\left(2+\sqrt{3}-k\right)$.

$$\begin{gathered} \sqrt{3}+\sqrt{3}(2+\sqrt{3})k=2+\sqrt{3}-k \\ \Rightarrow \sqrt{3}(2+\sqrt{3})k+k=2 \\ \Rightarrow \left(\sqrt{3}(2+\sqrt{3})+1\right)k=2 \\ \Rightarrow k=\frac{2}{\left(\sqrt{3}(2+\sqrt{3})+1\right)} \\ \Rightarrow k=\frac{2}{\left(2\sqrt{3}+4\right)} \end{gathered}$$

When the left-hand side is equal to $-\left(2+\sqrt{3}-k\right)$.

$$\begin{gathered} \sqrt{3}+\sqrt{3}(2+\sqrt{3})k=-2-\sqrt{3}+k \\ \Rightarrow \sqrt{3}(2+\sqrt{3})k-k=-2-2\sqrt{3} \\ \Rightarrow \left(\sqrt{3}(2+\sqrt{3})-1\right)k=-2\left(1+\sqrt{3}\right) \\ \Rightarrow k=\frac{-2\left(1+\sqrt{3}\right)}{\left(\sqrt{3}(2+\sqrt{3})-1\right)} \\ \Rightarrow k=\frac{-2\left(1+\sqrt{3}\right)}{2\left(\sqrt{3}+1\right)} \\ \Rightarrow k=-1 \end{gathered}$$

Therefore, the values of $k$ are $\frac{2}{2\sqrt{3}+4}$ and $-1$.

Hence, option (C) is the correct answer.