+129894 The x, y coordinates of F' =
[1cos(180) - 5sin(180), 1sin(180) + 5cos(180) ] = [ -1 , - 5 ]
And the x , y coordinates of G' =
[4cos(180) - (-3)sin(180), 4sin(180) + (-3)cos(180) ] = [ -4 , 3 ]
"b" is correct
+26393 confused
Matrix Rotation counterclockwise:
\(\begin{array}{|rcll|} \hline \begin{pmatrix} \cos(\varphi) & \sin (\varphi) \\ -\sin(\varphi) & \cos (\varphi) \\ \end{pmatrix} \stackrel{\varphi=180^{\circ}} \rightarrow \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix} \\ \hline \end{array} \)
The point P becomes to P':
\(\begin{array}{|rcll|} \hline \binom{x}{y}\cdot \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix} = \binom{-x}{-y} \\ \hline \end{array} \)
\(\text{Let}\ F =\binom{1}{5} \\ \text{Let}\ G =\binom{4}{-3} \)
\(\begin{array}{|rcll|} \hline F'=\binom{1}{5}\cdot \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix} = \binom{-1}{-5} \\ G'=\binom{4}{-3}\cdot \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix} = \binom{-4}{3} \\ \hline \end{array}\)
The answer is b.
