+0  
 
0
204
4
avatar+282 

i hate this

lizagame  Feb 28, 2017
 #1
avatar
0

each quadrant is 90 degrees, so you will turn it 3 times, because 90 * 3 = 270

Guest Feb 28, 2017
 #2
avatar+87301 
0

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

 

 

cool cool cool

CPhill  Feb 28, 2017
 #3
avatar+92781 
0

Try watching this.  There are a lot of resourced on the web to help with mathematical concepts.

If you want us to help you find them then ask specifically. :)

 

https://www.youtube.com/watch?v=5trp4U4rmLM

Melody  Feb 28, 2017
 #4
avatar+19632 
+5

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.

 

laugh

heureka  Mar 1, 2017

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