Question

The equations of the lines which pass through the point $\left(3,-2\right)$ and are inclined at $60°$ to the line $\sqrt{3}x+y=1$ are

• A. $y+2=0,\sqrt{3}x-y-2-3\sqrt{3}=0$
• B. $x-2=0,\sqrt{3}x-y+2+3\sqrt{3}=0$
• C. $\sqrt{3}x-y-2-3\sqrt{3}=0$
• D. None of the above

Answer 1

The correct option is A

$y+2=0,\sqrt{3}x-y-2-3\sqrt{3}=0$

Calculate the equations:

The given equation of line is $\sqrt{3}x+y=1\Rightarrow y=-\sqrt{3}x+1$

This is of the form $y=mx+c$

Slope of the line is $-\sqrt{3}$.

Let m be the slope of other line which make $60°$ with the given line.

We know $\tan \theta =\frac{m_2-m_1}{1+m_1{m}_2}$

$\begin{array}{rcl}\tan 60°& =& \frac{\left|\left(-\sqrt{3}-m\right)\right|}{\left|\left(1-\sqrt{3}m\right)\right|}\\ \sqrt{3}& =& \frac{\left|\left(-\sqrt{3}-m\right)\right|}{\left|\left(1-\sqrt{3}m\right)\right|}\\ \sqrt{3}-3m& =& -\sqrt{3}-mor\sqrt{3}-3m=\sqrt{3}+m\\ & \Rightarrow & m=\sqrt{3}orm=0\end{array}$

The line passes through $\left(3,-2\right)$.

Required equation is $\left(y-y_1\right)=m\left(x-x_1\right)$

$$\begin{gathered} \Rightarrow y+2=\sqrt{3}\left(x-3\right)\&y+2=0 \\ \Rightarrow \sqrt{3}x-y-2-3\sqrt{3}=0\&y+2=0 \end{gathered}$$

Hence $y+2=0,\sqrt{3}x-y-2-3\sqrt{3}=0$ are the required equations and therefore option $\left(A\right)$ is the answer.