How do you convert #5x-3y=6# to polar form?

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Answer 1 · 2016-05-24T14:31:53

#r cos(theta+arctan(3/5))=6/sqrt(34)#

Substituting the pass equations

#((x=r cos(theta)) ,( y=r sin(theta)))#

we get

#5r cos(theta)-3r sin(theta)=6#

Solving

#((5=r_0 cos(theta_0)), (-3 = r_0 sin(theta_0)))#

for #r_0, theta_0# we obtain #r_0=sqrt(34),a=-arctan(3/5)#
so the equivalence

#5r cos(theta)-3r sin(theta)=6 equiv r r_0(cos(theta_0)cos(theta)+sin(theta_0)sin(theta))=6#

can be simplified to

#r cos(theta-theta_0)=6/r_0#. Substituting the found values we get
#r cos(theta+arctan(3/5))=6/sqrt(34)#