How do you convert #(-8,0)# into polar forms?

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Answer 1 · 2018-05-05T18:28:55

#(8, pi)# (radians) or #(8, 180^@)# (degrees)

Rectangular #-># Polar: #(x, y) -> (r, theta)#

Find #r# (radius) using #r = sqrt(x^2 + y^2)#
Find #theta# by finding the reference angle: #tantheta = y/x# and use this to find the angle in the correct quadrant

#r = sqrt((-8)^2 + (0)^2)#

#r = sqrt(64)#

#r = 8#

Now we find the value of #theta# using #tantheta = y/x#.

#tantheta = 0/-8#

#tantheta = 0#

#theta = tan^-1(0)#

#theta = 0 or pi#

To determine which one it is, we have to look at our coordinate #(-8, 0)#. First, let's graph it:
enter image source here

As you can see, it is on the negative side of the #x# axis. Our #theta# has to match where that is, meaning that #theta = pi#.

From #r# and #theta#, we can write our polar coordinate:
#(8, pi)# (radians) or #(8, 180^@)# (degrees)

Hope this helps!