Question

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

Answer 1

To convert from rectangular to polar we will use these 2 formulas:

$r^2=x^2+y^2$ and

$tan\theta=\frac{y}{x}$.

The r value found serves as the first coordinate in our polar coordinate, and the angle serves as the second coordinate of the pair. We are told to find 2. Since the r value will always be the same (it's the length of the hypotenuse created in the right triangle we form when determining our angle theta), the angle is what is going to be different in our coordinate pairs. We use the x and y coordinates from the given rectangular coordinate to solve for the r in both our coordinate pairs.

$r^2=2^2+ -2^2$ which gives us an r value of $2\sqrt{2}$. That's r for both coordinate pairs. Now we move to the angle. Setting up according to our formula we have

$tan\theta =\frac{-2}{2} =-1$.

This asks the question "what angle(s) has/have a tangent of -1?". That's what we have to find out! Since the tangent ratio is y/x AND since it is negative, it is going to lie in a quadrant where x is negative and y is positive, AND where x is positive and y is negative. Those quadrants are 2 and 4. In QII, x is negative so the tangent ratio is negative here; in QIV, y is negative so the tangent ratio is negative here as well. Now, if we type inverse tangent of -1 into our calculators in degree mode, we get that the angle that has a tangent of -1 is -45. Measured from the positive x axis, -45 does in fact go into the fourth quadrant. However, since the inverse tangent of -1 is -45, we also have a 45 degree angle in the second quadrant. Those are reference angles, mind you. A 45 degree angle in QII has a coterminal angle of 135 degree; a 45 degree angle in QIV has a coterminal angle of 315. If you don't understand that, go back to your lesson on reference angles and coterminal angles to see what those are. So our polar coordinates for that rectangular coordinate are

$(2\sqrt{2},135)$ and

$(2\sqrt{2},315)$