Match the following.

Column - I		Column - II
I.	The Point $A\left(\sin \theta ,\cos \theta \right)$ is 3 units away from the point  $\left(\text{2cos75}°,\text{2sin75}°\right)$ If $0°<\theta <360°$ then $\sin \left(\theta -45°\right)=$	P.	$\frac{1}{2}$
II.	The harmonic conjugate of $\left(4,1\right)$with respect to the points $\left(3,2\right)$ and $\left(-1,6\right)$ is $\left(\alpha ,\beta \right)$ then $\alpha +\beta =$	Q.	$5$
III.	In a triangle ABC angle B is right angled,$\text{AC}=\text{2},\text{ A}\left(\text{2},\text{ 2}\right),\text{ B}\left(\text{1},\text{ 3}\right)$ then length of median   $\text{AD}=$	R.	$\sqrt{\frac{5}{2}}$

(1) $A=\left(\sin \theta ,\cos \theta \right)B=\left(2\cos 75°,2\sin 75°\right)$

$Put\theta =195°\sin 195°=-\sin 15°=-\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)$

$\cos 195°=-\cos 15°=-\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)$

$A=\left[\frac{-\left(\sqrt{3}-1\right)}{2\sqrt{2}},\frac{-\left(\sqrt{3}+1\right)}{2\sqrt{2}}\right]B\left[\frac{2\left(\sqrt{3}-1\right)}{2\sqrt{2}},\frac{2\left(\sqrt{3}+1\right)}{2\sqrt{2}}\right]$

$=\sqrt{{\left[\frac{\left(2\sqrt{3}-2+\sqrt{3}-1\right)}{2\sqrt{2}}\right]}^2+{\left[\frac{\left(2\sqrt{3}+2+\sqrt{3}+1\right)}{2\sqrt{2}}\right]}^2}$

$=\sqrt{\frac{{\left(3\sqrt{3}-1\right)}^2+{\left(3\sqrt{3}+3\right)}^2}{8}}$

$=\sqrt{\frac{2\left(27+9\right)}{8}}=3$

$\left(4,1\right)=\left[\frac{-\lambda +3}{\lambda +1},\frac{6\lambda +2}{\lambda +1}\right]$

$\therefore 3-\lambda =4\lambda +4and6\lambda +2=\lambda +1$

$5\lambda =-1\therefore \lambda =-\frac{1}{5}$

Conjugate ratio $=\text{ 1}:\text{5}$

Conjugate point = $\left[\frac{1\left(-1\right)+5\left(3\right)}{1+5},\frac{1\left(6\right)+5\left(2\right)}{1+5}\right]=\left[\frac{7}{3},\frac{8}{3}\right]$

$\text{ Since }AC=2,BA=\sqrt{2}$

$\therefore BC=\sqrt{2}$

$\therefore a=\sqrt{2},b=2,c=\sqrt{2}$

Length of median $AD=\frac{1}{2}\sqrt{2b^2+2c^2-a^2}$

$=\frac{1}{2}\sqrt{8+4-2}=\sqrt{\frac{10}{2}}=\sqrt{\frac{10}{4}}=\sqrt{\frac{5}{2}}$