How do you convert #r = 6 / (2cos[theta] - 3sin[theta]) # into cartesian form?

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Answer 1 · 2018-04-28T11:55:14

#2x - 3y=6 #

We know our polar conversions:

#r^2 = x^2 + y^2 #

#rcostheta = x #

#rsintheta = y #

#r = 6/ (2costheta - 3sintheta) #

#=> 2rcostheta - 3rsintheta = 6 #

#=>color(red)( 2x - 3y = 6 #

How do you convert #r = 6 / (2*cos[theta] - 3*sin[theta]) # into cartesian form?