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 ${15}^{}$. 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 equation of line in new position is
$y-0=\tan {15}^{}(x-2)$ ($\because$ initial $\theta ={30}^{}$, after rotating it becomes $\theta ={15}^{}$)
$\Rightarrow y=\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)(x-2)$
$\Rightarrow y=\frac{{(\sqrt{3}-1)}^2}{2}(x-2)$
$\Rightarrow 2y=(4-2\sqrt{3})(x-2)$
$\Rightarrow (2-\sqrt{3})x-y-4+2\sqrt{3}=0$