Question

The line joining two points $A(2,0),B(2+\sqrt{3},1)$ is rotated about $A$ in anti-clockwise direction through an angle of ${30}^{\circ }$. If the coordinates of new position of $B$ is $(h,k)$, then the value of $h^2+k^2$ is

Answer 1

Distance $AB=\sqrt{(\sqrt{3}{)}^2+1^2}=2\text{units}$
The slope of the line $AB$ is given by
$m=\frac{1-0}{2+\sqrt{3}-2}=\frac{1}{\sqrt{3}}$
$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}\Rightarrow \theta =\frac{\pi }{6}$
$\therefore$ The line makes on angle of ${30}^{\circ }$ with $x-$axis.
$AB$ is rotated through ${30}^{\circ }$about point $A$ in anticlock wise. So, it makes an angle of ${60}^{\circ }$ with $x-$axis.
The new coordinates of $B$ is
$(2+AB\cos \theta ,0+AB\sin \theta )=(2+2\cos {60}^{\circ },2\sin {60}^{\circ })=(3,\sqrt{3})\therefore h^2+k^2=9+3=12$