Question

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

Answer 1

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

Explanation:

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)`