How do you convert #x/y=8-x# into polar form? Trigonometry The Polar System Converting Between Systems 1 Answer Cesareo R. May 13, 2016 #r + (cot(theta)-8)sec(theta)=0# Explanation: The pass equations #x = r cos(theta), y = r sin(theta)# substituted into the cartesian equation gives: #(r cos(theta))/(r sin(theta)) = 8 - r cos(theta)#. Simplifiying and solving for #r# #r + (cot(theta)-8)sec(theta)=0# Answer link Related questions How do you convert rectangular coordinates to polar coordinates? When is it easier to use the polar form of an equation or a rectangular form of an equation? How do you write #r = 4 \cos \theta # into rectangular form? What is the rectangular form of #r = 3 \csc \theta #? What is the polar form of # x^2 + y^2 = 2x#? How do you convert #r \sin^2 \theta =3 \cos \theta# into rectangular form? How do you convert from 300 degrees to radians? How do you convert the polar equation #10 sin(θ)# to the rectangular form? How do you convert the rectangular equation to polar form x=4? How do you find the cartesian graph of #r cos(θ) = 9#? See all questions in Converting Between Systems Impact of this question 2395 views around the world You can reuse this answer Creative Commons License