Question

The line joining two points $A$ $(2,0)$ and $B(3,1)$ is rotated about $A$ in anticlockwise direction through an angle ${15}^o$. lf $B$ goes to $C$, then $C=$

• A. $\left(\frac{4+\sqrt{2}}{2},\sqrt{6}\right)$
• B. $\left(\frac{6+\sqrt{2}}{2},\frac{\sqrt{6}}{2}\right)$
• C. $\left(\frac{2+\sqrt{2}}{2},\frac{\sqrt{6}}{2}\right)$
• D. $\left(\frac{4+\sqrt{2}}{2},\frac{\sqrt{6}}{2}\right)$

Answer 1

Answer: (D)

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

Slope of line $AB=1$
$\Rightarrow \tan \theta =1$
So, $AB$ makes an angle of ${45}^0$ with the $x$-axis.
Now point $A$ is at $x$-axis only.
So, $AC$ makes an angle of ${60}^0$ with $x$-axis.
$AB=r=\sqrt{2}$
Given co-ordinates of $B$ are $(3,1)$
Let the co-ordinates of $C$ be $(x,y)$
$x=2+\sqrt{2}\cos {60}^0$

$x=2+\frac{1}{\sqrt{2}}=\frac{4+\sqrt{2}}{2}$

$y=0+\sqrt{2}\sin {60}^0$

$y=\frac{\sqrt{6}}{2}$