Question

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

• 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. $x+\sqrt{3}y-2=0$

Answer 1

The correct option is A $\sqrt{3}x-y-2\sqrt{3}=0$

Slope of $AB=\frac{1}{1}$

$\Rightarrow \tan \theta =m_1=1$ or $\theta ={45}^{\circ }$

Thus, slope of new line is $\tan ({45}^{\circ }+{15}^{\circ })=\tan {60}^{\circ }=\sqrt{3}$

$(\because$ 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 $y=\sqrt{3}x-2\sqrt{3}$

$⟹\sqrt{3}x-y-2\sqrt{3}=0$