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The line joining two points $A (2, 0), B (3, 1)$ is rotated about $A$ in anticlockwise direction through an angle $15 °$ .If $B$ goes to $C$ then $C =$

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$(\frac{4 + \sqrt{2}}{2}, \frac{\sqrt{6}}{2})$

$(\frac{4 + \sqrt{2}}{2}, \sqrt{6})$

$(\frac{6 + \sqrt{2}}{2}, \frac{\sqrt{6}}{2})$

$(\frac{2 + \sqrt{2}}{2}, \frac{\sqrt{6}}{2})$

answer is D.

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Detailed Solution

$A = (2, 0), B (3, 1)$ Given angle $= 15 °$

$m = 1 \to θ = 45 °$
Total angle $= 45 ° + 15 ° = 60 °$

$\begin{array}{l} A B = \sqrt{2} \\ C = (2 + \sqrt{2} . \cos 60 °, 0 + \sqrt{2} . \sin 60 °) \\ = (2 + \sqrt{2} . \frac{1}{2}, 0 + \sqrt{2} . \frac{\sqrt{3}}{2}) \\ = (\frac{4 + \sqrt{2}}{2}, \frac{\sqrt{6}}{2}) \end{array}$

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