If the line joining two points $\mathrm{A}(2,0)$ and $\mathrm{B}(3,1)$ is rotated about $\mathrm{A}$ in…
- $y=3 x-6$
- $y=\sqrt{3} x-2 \sqrt{3}$
- $y=-\sqrt{3} x+2 \sqrt{3}$
- $y=\frac{1}{\sqrt{3}} x-\frac{2}{\sqrt{3}}$
Solution
Slope of $\mathrm{AB}=\frac{1-0}{3-2}=1$
$\therefore \tan \theta=1 \Rightarrow \theta=45^{\circ}$
Line $\mathrm{AB}$ is rotated through $15^{\circ}$ in anticlockwise direction about
A.
Therefore in a new position, slope of line $=\tan \left(45^{\circ}+15^{\circ}\right)=\tan 60^{\circ}=\sqrt{3}$ and it passes through $\mathrm{A}$.
Required equation of line is $(\mathrm{y}-0)=\sqrt{3}(\mathrm{x}-2) \Rightarrow$
$\sqrt{3} x-2 \sqrt{3}=y$Asked in: MHT CET 2021 (22 Sep Shift 2)