Question

Determine two pairs of polar coordinates for the point (2, -2) with 0° ≤ θ < 360°.

answers

(2 square root of 2, 225°), (-2 square root of 2, 45°)

(2 square root of 2, 135°), (-2 square root of 2, 315°)

(2 square root of 2, 315°), (-2 square root of 2, 135°)

(2 square root of 2, 45°), (-2 square root of 2, 225°)

Answer 1

sit has been a bit laggy.. so

Answer 2

what? @jdoe0001

Answer 3

wooops I meant, the site gets laggy

Answer 4

as in the website

Answer 5

\(\large {
\begin{array}{cccllll}
(&2, &-2)\\
&\uparrow &\uparrow \\
&x&y
\end{array}\qquad
\begin{array}{llll}
r^2=x^2+y^2\to {\color{brown}{ r}}=\sqrt{x^2+y^2}
\\ \quad \\
tan(\theta)=\frac{y}{x}\to {\color{brown}{ \theta}}=tan^{-1}\left(\frac{y}{x}\right)
\end{array}
}\)

what would that give you for the "r" or radius of those coordinates and for the angle?

Answer 6

would the answer be (2 square root of 2, 45°), (-2 square root of 2, 225°)
@jdoe0001

Answer 7

hmmm what did you get for "r"?

Answer 8

\[\sqrt{8}\]

Answer 9

and... for \(\theta?\)

Answer 10

or (2 square root of 2, 225°), (-2 square root of 2, 45°)

Answer 11

for \[\theta \] 45

Answer 12

well.... the inverse tangent function, has a range of \(\large -\frac{\pi}{2}<\theta<\frac{\pi}{2}\) so the 45 degree is is just a reference angle

keep in mind that tangent of that angle, is a negative -1 since y/x = -1

and tangent is negative in only 2 places.... 2nd and 4th Quadrants

because in the 2nd Quadrant, x is negative and y positive

and in the 4th Quadrant, x is positive but y is negative