Question

A line passing through the origin and the point $A(\sqrt{3},1)$ is rotated through an angle of ${15}^{\circ }$ about the origin in anti-clockwise direction. If the point $B$ be the new position of $A$, then the product of the co-ordinates of $B$ is

Answer 1

$\text{Slope of the line,}OA=\frac{1-0}{\sqrt{3}-0}=\frac{1}{\sqrt{3}}\Rightarrow OA\text{makes an angle of}{30}^{\circ }.\text{So, the angle of new position is}{30}^{\circ }+{15}^{\circ }={45}^{\circ }\text{Since,}OB\text{passes through origin}.\therefore \text{Equation of}OB\text{is},\frac{x-0}{\cos {45}^{\circ }}=\frac{y-0}{\sin {45}^{\circ }}OA=\sqrt{(\sqrt{3}-0{)}^2+(1-0{)}^2}=2\therefore \text{The co-ordinates of}B\text{are given by,}\frac{x}{\frac{1}{\sqrt{2}}}=\frac{y}{\frac{1}{\sqrt{2}}}=2\Rightarrow x=\sqrt{2},y=\sqrt{2}\Rightarrow xy=\sqrt{2}\times \sqrt{2}=2$