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

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Answer 1 · 2016-06-09T09:41:08

$ρ sin (θ + 1.3734) - 0.784465 = 0$ $rho sin(theta+1.3734)-0.784465 = 0$

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

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

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