Question

If the line joining the points $A(3,0)$ and $B(5,2)$ is rotated about $A$ in the anticlockwise direction through an angle of ${15}^0$ such that $B$ goes to $C$ in the new position .Similarly the line joining $A$ and $C$ is rotated about $A$ in the anticlockwise direction through an angle of ${45}^{\circ }$ such that $C$ goes to $D$ in the new position, then the coordinates of $D$ are

• A. $(4-\sqrt{3},\sqrt{3}-1)$
• B. $(4+\sqrt{3},\sqrt{3}-1)$
• C. $(4-\sqrt{3},\sqrt{3}+1)$
• D. $(4+\sqrt{3},\sqrt{3}+1)$

Answer 1

The correct option is C $(4-\sqrt{3},\sqrt{3}+1)$



$m_{AB}=\frac{2}{2}=1$
$AB=AC=AD$
$=\sqrt{(5-3{)}^2+(2-0{)}^2}$
$=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$
Using parametric form,
$D=(3+r\cos \theta ,0+r\sin \theta );r=2\sqrt{2};\theta ={105}^{\circ }$
$D=(3+2\sqrt{2}\cos {105}^{\circ },2\sqrt{2}\sin {105}^{\circ })$
$=(3-2\sqrt{2}\sin {15}^{\circ },2\sqrt{2}\cos {15}^{\circ })$
$=(3-\frac{2\sqrt{2}(\sqrt{3}-1)}{2\sqrt{2}},2\sqrt{2}\frac{\sqrt{3}+1}{2\sqrt{2}})$
$\therefore D=(4-\sqrt{3},\sqrt{3}+1)$