If the point P(1,3) undergoes the following t | vedclass

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Question

(A) Step $1$: Reflection of $P(1,3)$ about the line $y=x$ gives $Q(3,1)$.Step $2$: Translation of $Q(3,1)$ by $3$ units along the positive $X$-axis gi

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Step $1$: Reflection of $P(1,3)$ about the line $y=x$ gives $Q(3,1)$.Step $2$: Translation of $Q(3,1)$ by $3$ units along the positive $X$-axis gives $R(3+3, 1) = R(6,1)$.Step $3$: Rotation of $R(6,1)$ by $	heta = \frac{\pi}{6}$ clockwise about the origin. The rotation matrix for clockwise rotation by $	heta$ is $\begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$.Here $	heta = \frac{\pi}{6}$,so $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin \frac{\pi}{6} = \frac{1}{2}$.The new coordinates $(x', y')$ are given by:$x' = 6 \cos \frac{\pi}{6} + 1 \sin \frac{\pi}{6} = 6 \left(\frac{\sqrt{3}}{2}ight) + 1 \left(\frac{1}{2}ight) = \frac{6 \sqrt{3}+1}{2}$$y' = -6 \sin \frac{\pi}{6} + 1 \cos \frac{\pi}{6} = -6 \left(\frac{1}{2}ight) + 1 \left(\frac{\sqrt{3}}{2}ight) = \frac{\sqrt{3}-6}{2}$Thus,the final position is $\left(\frac{6 \sqrt{3}+1}{2}, \frac{\sqrt{3}-6}{2}ight)$.

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