Polar Coordinate function of a Straight Line I was having some problem when trying to come out a polar coordinate function with straight line equation. 

I know it is not good to post images here, but please bear with me as the question requires us to solve the equation from the straight line in the image.

What I have done is I tried to come out with an implicit function for the straight line L.

2x + 3y - 6 = 0


But then I not sure how can I continue from there to come out with a polar coordinate function. Any ideas?

Thanks!
$x = r\cos \theta\\
y = r\sin\theta\\
2x + 3y - 6 = 0\\
2r\cos\theta + 3r\sin\theta = 6\\
r(2\cos\theta + 3\sin\theta) = 6$

Now I could say:

$r = \frac {6}{3\cos\theta + 2\sin\theta}$

and be done.

But I think that this is a little bit more informative:

$\sin (\arctan \frac ab) = \frac {a}{\sqrt {a^2 + b^2}}\\
\cos (\arctan \frac ab) = \frac {b}{\sqrt {a^2 + b^2}}$

$r\sqrt {2^2 + 3^2}(\sin (\arctan \frac{3}{2})\sin\theta + \cos(\arctan \frac 32)\cos\theta) = 6\\
r\sqrt {13}(\cos(\theta - \arctan \frac 32)) = 6\\
r\sqrt {13}(\cos(\theta - \arctan \frac 32)) = 6\\
r = \frac {6}{\sqrt {13}} \sec (\theta - \arctan \frac 32)$

As it gives the angle of rotation and the distance to the line.
Given the linear equation in two intercept form, x/a+y/b=1
The equivalent polar form is r=(absec(theta))/(b+tan(theta))