#1**+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

#1**+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