Rotating the figure 270° counter-clockwise is the same as rotating the figure 90° clockwise...
And each point will swap coordinates and both will have positive signs....so
(-2,1) → (1, 2)
(-4,2) → (2, 4)
(-3,5) → (5, 3)
C is correct
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Matrix Rotation counterclockwise:
\(\begin{array}{|rcll|} \hline \begin{pmatrix} \cos(\varphi) & \sin (\varphi) \\ -\sin(\varphi) & \cos (\varphi) \\ \end{pmatrix} \stackrel{\varphi=270^{\circ}} \rightarrow \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} \\ \hline \end{array} \)
The point P becomes to P':
\(\begin{array}{|rcll|} \hline \binom{x}{y}\cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} = \binom{y}{-x} \\ \hline \end{array} \)
\(\text{Let}\ X =\binom{-2}{1} \\ \text{Let}\ Y =\binom{-4}{2} \\ \text{Let}\ Z =\binom{-3}{5} \\ \)
\(\begin{array}{|rcll|} \hline X'=\binom{-2}{1}\cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} = \binom{1}{2} \\ Y'=\binom{-4}{2}\cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} = \binom{2}{4} \\ Z'=\binom{-3}{5}\cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} = \binom{5}{3} \\ \hline \end{array} \)
The answer is c.