Question

A line through the point $A(2,0)$ which makes an angle of $30°$with the positive direction of X-axis is rotated about$A$ in clockwise direction through an angle of $15°$ . Then, the equation of the straight line in the new position is

• A. $(2–\sqrt{3})x+y+4-2\sqrt{3}=0$
• B. $(2–\sqrt{3})x–y–4+2\sqrt{3}=0$
• C. $(2–\sqrt{3})x+y-4-2\sqrt{3}=0$
• D. $(2–\sqrt{3})x+y+4+2\sqrt{3}=0$

Answer 1

The correct option is B

$(2–\sqrt{3})x–y–4+2\sqrt{3}=0$

The explanation for the correct option:

Step1. Finding the slope of straight line.

Given,

A line through $A(2,0)$ makes $30°$with the positive direction of the x-axis.

When the line is rotated clockwise through $15°$ , the new angle is $15°$.

The general equation of a line is given by $y=mx+c$, where $m$ is slope $=\frac{y}{x}$

$=\tan \theta$

$\theta =15°$

Slope, $\begin{array}{rcl}m& =& \tan 15\\ & =& 2–\sqrt{3}\end{array}$

Step2. Finding the equation of line

General equation of line having slope through given point is given by,

$\mathit{y}\mathbf{-}{\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathit{m}\mathbf{(}\mathit{x}\mathbf{-}{\mathit{x}}_{\mathbf{1}}\mathbf{)}$

Equation of line with slope $m$ and passing through the point $A(2,0)$ is given by

$\begin{array}{rcl}y-0& =& (2-\sqrt{3})(x-2)\\ y& =& (2-\sqrt{3})(x-2)\\ \mathbf{(}\mathbf{2}\mathbf{-}\sqrt{\mathbf{3}}\mathbf{)}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{+}\mathbf{2}\sqrt{\mathbf{3}}& \mathbf{=}& \mathbf{0}\end{array}$

Hence, option $\mathbf{\left(}\mathit{B}\mathbf{\right)}$ is the correct answer.