First slide

Cartesian plane

Question

If a line joining two points $A (2, 0)$ and $B (3, 1)$ is rotated about $A$ in anticlockwise direction through an angle $15 °$ such that the point $B$ goes to $C$ in the new position, then the coordinates of $C$ are

Moderate

A

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

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

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

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

Solution

The equation of $A C$ is
$\frac{x - 2}{\cos 60^{\circ}} = \frac{y - 0}{\sin 60^{\circ}} = r$
We have, $A B = \sqrt{(3 - 2)^{2} + (1 - 0)^{2}} = \sqrt{2}$
As $A C$ is the new position of $A B .$ Therefore, $A C = A B = \sqrt{2}$ .
Thus, the coordinates of $C$ are given by
$\frac{x - 2}{\cos 60^{\circ}} = \frac{y - 0}{\sin 60^{\circ}} = \sqrt{2} \Rightarrow x = 2 + \frac{1}{\sqrt{2}}, y = \sqrt{\frac{3}{2}}$
Hence, the coordinates of $C$ are $(2 + \frac{1}{\sqrt{2}}, \sqrt{\frac{3}{2}}) .$

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