How do you convert #(11,-9)# into polar coordinates? Precalculus Polar Coordinates Converting Coordinates from Polar to Rectangular 1 Answer 1s2s2p May 6, 2018 #(sqrt202,tan^-1(-9/11)+2pi) or (14.2,5.60^c)# Explanation: #(x,y)->(r,theta);(r,theta)=(sqrt(x^2+y^2),tan^-1(y/x))# #r=sqrt(x^2+y^2)=sqrt(11^2+(-9)^2)=sqrt(121+81)=sqrt202~~14.2# #theta=tan^-1(-9/11)# However, #(11,-9)# is in quadrant 4, and so we must add #2pi# to our answer. #theta=tan^-1(-9/11)+2pi ~~5.60^c# #(sqrt202,tan^-1(-9/11)+2pi) or (14.2,5.60^c)# Answer link Related questions What is the formula for converting polar coordinates to rectangular coordinates? How do I convert polar coordinates #(5, 30^circ)# to rectangular coordinates? How do I convert polar coordinates #(3.6, 56.31)# to rectangular coordinates? How do I convert polar coordinates #(10, -pi/4)# to rectangular coordinates? How do I convert polar coordinates #(4,-pi/3)# to rectangular coordinates? How do I convert polar coordinates #(6, 60^circ)# to rectangular coordinates? How do I convert polar coordinates #(-4, 230^circ)# to rectangular coordinates? What is the Cartesian equivalent of polar coordinates #(sqrt97, 66^circ)#? What is the Cartesian equivalent of polar coordinates #(2, pi/6)#? What is the Cartesian equivalent of polar coordinates #(7, pi)#? See all questions in Converting Coordinates from Polar to Rectangular Impact of this question 1834 views around the world You can reuse this answer Creative Commons License