Q. A line through the point $A(2,\ 0)$ which makes an angle of $30{^\circ}$ with the positive direction of $x$-axis is rotated about $A$ in clockwise direction through an angle of $15{^\circ}$. Then, the equation of the straight line in the new position is
A) $\left( 2 - \sqrt{3}\ \right)x + y - 4 + 2\sqrt{3} = 0$
B) $\left( 2 - \sqrt{3} \right)x - y - 4 + 2\sqrt{3} = 0$
C) $\left( 2 - \sqrt{3} \right)x - y + 4 + 2\sqrt{3} = 0$
D) $\left( 2 - \sqrt{3} \right)x + y + 4 + 2\sqrt{3} = 0$
Explanation
(b)
The equation of line in new position is
$y - 0 = \tan{15{^\circ}}(x - 2)$
$\Rightarrow \ y = \left( 2 - \sqrt{3} \right)(x - 2)$
$\Rightarrow \ \left( 2 - \sqrt{3} \right)x - y - 4 + 2\sqrt{3} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.