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A pre-image point is rotated 90° clockwise. If the pre-image point had the coordinates (3, -5), what are the coordinates of its image point
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Much easier to find with graph paper.
\[ \Large \left(\begin{matrix}x' \\ y'\end{matrix}\right)=\left(\begin{matrix}\cos \alpha & - \sin \alpha \\ \sin \alpha & \cos \alpha\end{matrix}\right)\left(\begin{matrix}x \\ y\end{matrix}\right)\]
not sure if you know how to apply that already, however that's for a matrix projection
You do not need to use that to solve this problem.
For rotation around the origin with \(\alpha = 90^\circ \) the points just become normal to each other, so components exchange and signs flip for one, in this case (-5/3)
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