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SOLUTION??

 Jun 8, 2019

Best Answer 

 #1
avatar+8853 
+4

\((2,3)\)  is at the polar coordinates of  \((r,\theta)\)  so:

 

\(2\ =\ r\cos\theta\\~\\ 3\ =\ r\sin\theta\)

 

 

\((x_1,y_1)\)  is at the polar coordinates of  \((2r,\theta+\frac{\pi}{2})\)  so:

 

\(\begin{array}{} x_1\ =\ (2r)\cos(\theta+\frac{\pi}{2})&=&(2r)(-\sin\theta)&=&(-2)(r\sin\theta)&=&(-2)(3)&=&-6\\~\\ y_1\ =\ (2r)\sin(\theta+\frac{\pi}{2})&=&(2r)(\cos\theta)&=&(2)(r\cos\theta)&=&(2)(2)&=&4 \end{array}\)

 

 

\((x_2,y_2)\)  is at the polar coordinates of  \((-r,-\theta)\)  so:

 

\(\begin{array}{c} x_2\ =\ (-r)\cos(-\theta)&=&(-r)\cos\theta&=&-(r\cos\theta)&=&-2\\~\\ y_2\ =\ (-r)\sin(-\theta)&=&(-r)(-\sin\theta)&=&r\sin\theta&=&3 \end{array}\)_

 Jun 9, 2019
 #1
avatar+8853 
+4
Best Answer

\((2,3)\)  is at the polar coordinates of  \((r,\theta)\)  so:

 

\(2\ =\ r\cos\theta\\~\\ 3\ =\ r\sin\theta\)

 

 

\((x_1,y_1)\)  is at the polar coordinates of  \((2r,\theta+\frac{\pi}{2})\)  so:

 

\(\begin{array}{} x_1\ =\ (2r)\cos(\theta+\frac{\pi}{2})&=&(2r)(-\sin\theta)&=&(-2)(r\sin\theta)&=&(-2)(3)&=&-6\\~\\ y_1\ =\ (2r)\sin(\theta+\frac{\pi}{2})&=&(2r)(\cos\theta)&=&(2)(r\cos\theta)&=&(2)(2)&=&4 \end{array}\)

 

 

\((x_2,y_2)\)  is at the polar coordinates of  \((-r,-\theta)\)  so:

 

\(\begin{array}{c} x_2\ =\ (-r)\cos(-\theta)&=&(-r)\cos\theta&=&-(r\cos\theta)&=&-2\\~\\ y_2\ =\ (-r)\sin(-\theta)&=&(-r)(-\sin\theta)&=&r\sin\theta&=&3 \end{array}\)_

hectictar Jun 9, 2019

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