The line joining two points A (2, 0), B( 3, 1) is rotated about A in anti clock wise direction through an angle 15⁰. If B goes to C then C=

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$(2 + \frac{1}{\sqrt{2}}, \sqrt{\frac{3}{2}})$

$(2 + \frac{1}{\sqrt{2}}, \frac{\sqrt{3}}{2})$

$(\sqrt{2} - 1, \frac{\sqrt{3}}{2})$

$(\sqrt{2} - \frac{1}{2}, \frac{\sqrt{2}}{3})$

answer is A.

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Detailed Solution

Given A (2, 0) B ( 3, 1)

Slope of $A B = \frac{1 - 0}{3 - 2} = 1$

Angle made by the line AB with X- axis is $\frac{π}{4}$

$\begin{array}{l} ∴ θ = A n g l e m a d e b y t h e l i n e \bar{A C} w i t h X - a x i s \\ = θ = 45^{0} + 15^{0} = 60^{0} \\ r = A C = A B = \sqrt{{(3 - 2)}^{2} + {(1 - 0)}^{2}} = \sqrt{1 + 1} = \sqrt{2} \\ ∴ C = (x_{1} + r \cos θ y_{1} + r \sin θ) \\ = (2 + \sqrt{2} \cos 60^{0} 0 + \sqrt{2} \sin 60^{0}) \\ = (2 + \sqrt{2} (\frac{1}{2}) \sqrt{2} (\frac{\sqrt{3}}{2})) \\ = (2 + \frac{1}{\sqrt{2}}, \sqrt{\frac{3}{2}}) \end{array}$