How do you convert #r= 1/(6cosθ-2sinθ)# into cartesian form?

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Answer 1 · 2016-11-21T09:09:13

#6x-2y=1#

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#