Question

The line joining two points $ A(2, 0) $ and $ B(3, 1) $ is rotated about $ A $ in an anticlockwise direction through an angle of $ 15^\circ $. If $ B $ goes to $ C $ in the new position, then the coordinates of $ C $ are:

• A. \( \left( 2 + \frac{1}{\sqrt{3}} \sqrt{3} , 2 \right) \)
• B. \( \left( 2 , \sqrt{\frac{3}{2}} \right) \)
• C. \( \left( 2 + \frac{1}{\sqrt{3}} , 1 \right) \)
• D. \( \left( 2 + \frac{1}{\sqrt{2}} , \sqrt{3} \right) \)

Hint:

When rotating points, use the rotation formulas for \( x' \) and \( y' \) to get the new coordinates. Ensure to use the correct angle in radians or degrees and check the direction of rotation (anticlockwise or clockwise).

Answer 1

The Correct Option is D

To find the new coordinates after rotating the point \( B(3, 1) \) about the point \( A(2, 0) \) by an angle \( \theta = 15^\circ \), we use the rotation formula: \[ x' = x_0 + (x - x_0) \cos \theta - (y - y_0) \sin \theta \] \[ y' = y_0 + (x - x_0) \sin \theta + (y - y_0) \cos \theta \] Here, \( A(2, 0) \) is the center of rotation and \( B(3, 1) \) is the point being rotated. We first compute the differences \( x - x_0 \) and \( y - y_0 \): \[ x - x_0 = 3 - 2 = 1 \quad \text{and} \quad y - y_0 = 1 - 0 = 1 \] Now, substituting into the rotation formula for \( \theta = 15^\circ \): \[ x' = 2 + (1) \cos 15^\circ - (1) \sin 15^\circ = 2 + \frac{1}{\sqrt{2}} = 2 + \frac{1}{\sqrt{3}} \] \[ y' = 0 + (1) \sin 15^\circ + (1) \cos 15^\circ = \sqrt{3} \] Thus, the coordinates of \( C \) are: \[ C = \left( 2 + \frac{1}{\sqrt{2}}, \sqrt{3} \right) \]