We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# PLEASE HELP THANKS!!!

0
44
1

SOLUTION??

Jun 8, 2019

### Best Answer

#1
+8215
+3

$$(2,3)$$  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+0 Answers

#1
+8215
+3
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