How do you convert #y=-5x+4# to polar form?

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Answer 1 · 2016-06-09T09:41:08

#rho sin(theta+1.3734)-0.784465 = 0#

The pass equations are

#{ (x = rho cos(theta)), (y = rho sin(theta)) :}#

substituting in

#y + a x + b=0#

we get

#rho (sin(theta)+a cos(theta))+b=0#

considering now #a = tan(theta_0)# and substituting

#rho(sin(theta)cos(theta_0)+cos(theta)sin(theta_0))+b cos(theta_0)=0#

but

#sin(alpha + beta) = Cos(beta) Sin(alpha) + Cos(alpha)Sin(beta)#

then the short version is

#rho sin(theta+theta_0)+bcos(theta_0) = 0#

in the present case

#theta_0=arctan(5)=1.3734# and #cos(theta_0)=0.196116#

so the polar formulation is

#rho sin(theta+1.3734)-0.784465 = 0#