The line joining two points $A(2,0), B(3,1)$ is rotated about $A$ in anti-clockwise direction through an angle of $15^{\circ}$. The equation of the line in the now position, is

• Correct answer
  A

  $\sqrt{3} x-y-2 \sqrt{3}=0$
• B

  $x-3 \sqrt{y}-2=0$
• C

  $\sqrt{3} x+y-2 \sqrt{3}=0$
• D

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

Solution

$\begin{array}{l} \text { Slope of } \mathrm{AB}=\frac{1}{1} \\ \Rightarrow \tan \theta=\mathrm{m}_{1}=1 \text { or } \theta=45^{\circ} \end{array}$ Thus, slope of new line is $\tan \left(45^{\circ}+15^{\circ}\right)$ $=\tan 60^{\circ}=\sqrt{3}$ $(\because$ it is rotated anti-clockwise, so the angle will be $45^{\circ}+15^{\circ}=60^{\circ}$ )
Hence, the equation is $y=\sqrt{3} x+c$ But it passes through $(2,0)$ So, $c=-2 \sqrt{3}$ Thus, required equation is $y=\sqrt{3} x-2 \sqrt{3}$

Hence, the correct answer is option: 'A'.