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!


If  $\alpha $ is CCW angle made to the normal then in Cartesian form ( not exactly as shown in your question, it should be normal to the line L.
$$ x \cos\alpha + y \sin  \alpha = p $$
and in polar coordinates after transformation
$$ r = p \sec \left( \theta - \tan^{-1} \frac{a}{b} \right ) $$
