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

Answer 1

Answer: (A)

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

Answer 2

Here, slope of $A B=\frac{1}{1}$
$\Rightarrow \tan \theta=m_{1}=1$ or $\theta=45^{\circ}$
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 $\left.45^{\circ}+15^{\circ}=60^{\circ}\right)$

Hence, the equation is $y=\sqrt{3} x+c$,
but it still passes through $(2,0)$,
hence $c=-2 \sqrt{3}$.
Thus, required equation is
$y=\sqrt{3} x-2 \sqrt{3}$