How do you convert #r= 1/(6cosθ-2sinθ)# into cartesian form? Precalculus Polar Coordinates Converting Equations from Polar to Rectangular 1 Answer sjc Nov 21, 2016 #6x-2y=1# Explanation: for polar co-ordinates #r^2=x^2+y^2# #color(red)(x=rcostheta)# #color(blue)(y=rsintheta)# for#" "r=1/(6costheta-2sintheta)# rearrange #r(6costheta-2sintheta)=1# #6color(red)(rcostheta)-2color(blue)(rsintheta)=1# #6color(red)(x)-2color(blue)(y)=1# Answer link Related questions What is the polar equation of a horizontal line? What is the polar equation for #x^2+y^2=9#? How do I graph a polar equation? How do I find the polar equation for #y = 5#? What is a polar equation? How do I find the polar equation for #x^2+y^2=7y#? How do I convert the polar equation #r=10# to its Cartesian equivalent? How do I convert the polar equation #r=10 sin theta# to its Cartesian equivalent? How do you convert polar equations to rectangular equations? How do you convert #r=6cosθ# into a cartesian equation? See all questions in Converting Equations from Polar to Rectangular Impact of this question 2072 views around the world You can reuse this answer Creative Commons License