Question

The point $P(1,3)$ undergoes the following transformations successively:
(i) Reflection with respect to the line $y=x$
(ii) Translation through $3$ units along the positive of the $x$-axis
(iii) Rotation through an angle of $\frac{\pi }{6}$ about the origin in the clockwise direction.
The final position of the point $P$ is

• A. $(\frac{6\sqrt{3}+1}{2},\frac{\sqrt{3}-6}{2})$
• B. $(\frac{7}{\sqrt{2}},\frac{-5}{\sqrt{2}})$
• C. $(\frac{6+\sqrt{3}}{2},\frac{1-6\sqrt{3}}{2})$
• D. $(\frac{6\sqrt{3}-1}{2},\frac{6+\sqrt{3}}{2})$

Answer 1

The correct option is A $(\frac{6\sqrt{3}+1}{2},\frac{\sqrt{3}-6}{2})$

i) After reflection, new position $P_1=(3,1)$

ii) After translation, new position $P_2=(3+3,1)=(6,1)$

iii) After rotation, new position $P_3$ will be $\left(r\cos \right(Q-\frac{\pi }{6}),r\sin (Q-\frac{\pi }{6}))$

Converting $P_2$ in polar form

$r=\sqrt{6^2+1^2}=\sqrt{37}$

$\tan Q=\frac{1}{6},\cos Q=\frac{6}{r},\sin Q=\frac{1}{r}$

$\therefore$ New trasn. $P_3=(r\cos (Q-\frac{\pi }{6}),r\sin (Q-\frac{\pi }{6}))$

$=\left(r\right[\cos Q\cos \frac{\pi }{6}+\sin Q\sin \frac{\pi }{6})$,$r(\sin Q\cos \frac{\pi }{6}-\cos Q\sin \frac{\pi }{6})$

$=(r\cdot \frac{6}{r}\cdot \frac{\sqrt{3}}{2}+r\cdot \frac{1}{r}\cdot \frac{1}{2},r\cdot \frac{1}{r}\cdot \frac{\sqrt{3}}{2}-r\cdot \frac{6}{r}\cdot \frac{1}{2})$

$=(\frac{6\sqrt{3}+1}{2},\frac{\sqrt{3}-6}{2})$