How do you find the polar equation for $3 x - 2 y = 1$ $3x-2y=1$ ?

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Answer 1 · 2016-11-08T18:44:10

The polar equation is $r = \frac{1}{3 cos θ - 2 sin θ}$ $r=1/(3costheta-2sintheta)$

We use, $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ , to go from cartesian to polar coordinates.

$3 x - 2 y = 1$ $3x-2y=1$ $\Rightarrow$ $=>$ $3 r cos θ - 2 r sin θ = 1$ $3rcostheta-2rsintheta=1$

$r = \frac{1}{3 cos θ - 2 sin θ}$ $r=1/(3costheta-2sintheta)$

Answer 2 · 2016-11-08T18:45:06

$ρ = \frac{1}{3 cos (θ) - 2 sin (θ)}$ $rho=1/(3cos(theta)-2sin(theta))$

$x = ρ cos (θ)$ $x=rhocos(theta)$
$y = ρ sin (θ)$ $y=rhosin(theta)$
$ρ = \sqrt[2]{x^{2} + y^{2}}$ $rho=root2(x^2+y^2)$

$3 ρ cos (θ) - 2 ρ sin (θ) = 1$ $3rhocos(theta)-2rhosin(theta)=1$
$ρ = \frac{1}{3 cos (θ) - 2 sin (θ)}$ $rho=1/(3cos(theta)-2sin(theta))$

How do you find the polar equation for 3x-2y=13x−2y=13x-2y=1?