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(A) The given point is $P(2, -1)$.The reflection of point $P(2, -1)$ about the $y$-axis is $P'(-2, -1)$.When the axes are rotated by an angle $	heta$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given point is $P(2, -1)$.The reflection of point $P(2, -1)$ about the $y$-axis is $P'(-2, -1)$.When the axes are rotated by an angle $	heta$ in the negative direction (clockwise),the rotation angle is $	heta = -45^{\circ}$.The transformation equations for rotation of axes are:$X = x \cos 	heta + y \sin 	heta$$Y = -x \sin 	heta + y \cos 	heta$Substituting $x = -2$,$y = -1$,and $	heta = -45^{\circ}$:$X = (-2) \cos(-45^{\circ}) + (-1) \sin(-45^{\circ}) = (-2) \left( \frac{1}{\sqrt{2}} ight) + (-1) \left( -\frac{1}{\sqrt{2}} ight) = -\frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$$Y = -(-2) \sin(-45^{\circ}) + (-1) \cos(-45^{\circ}) = 2 \left( -\frac{1}{\sqrt{2}} ight) - 1 \left( \frac{1}{\sqrt{2}} ight) = -\frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{3}{\sqrt{2}}$Thus,the new coordinates are $\left( -\frac{1}{\sqrt{2}}, -\frac{3}{\sqrt{2}} ight)$.

Find the new coordinates of the reflection of the point $(2, -1)$ about the $y$-axis,when the axes are rotated by $45^{\circ}$ in the negative direction without shifting the origin.

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