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SOLUTION??

Jun 8, 2019

#1
+8759
+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
+8759
+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:

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

hectictar Jun 9, 2019