Question

The transformed coordinates of the point $(4,3)$ when the axes are translated to point $(3,1)$ and then rotated through ${30}^{\circ }$ in anticlockwise direction is

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

Answer 1

The correct option is C $(\frac{\sqrt{3}+2}{2},\frac{2\sqrt{3}-1}{2})$

When $(0,0)$ is shifted to $(3,1)$
$\Rightarrow x=X+3,y=Y+1\Rightarrow (X,Y)=(4-3,3-1)=(1,2)$
Now, since coordinate axes are rotated by ${30}^{\circ }$, let coordinates of point be $(X^{\prime },Y^{\prime })$
$\Rightarrow X^{\prime }=X\cos {30}^{\circ }+Y\sin {30}^{\circ }$
And
$Y^{\prime }=-X\sin {30}^{\circ }+Y\cos {30}^{\circ }$

Therefore,
$X^{\prime }=1\cdot \frac{\sqrt{3}}{2}+2\cdot \frac{1}{2}=\frac{\sqrt{3}+2}{2}{Y}^{\prime }=-\frac{1}{2}+\frac{2\sqrt{3}}{2}=\frac{2\sqrt{3}-1}{2}$