Polar Coordinate function of a Straight Line

Question

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!

Answer 1

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

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.

Answer 2

In Cartesian coordinates, a straight line equation is $y=mx+b$ where is $m$ is a numerical slope and $b$ is a numerical $y$ intercept. Following rules for converting to polar coordinates, we let $x=r\cdot cos\theta$ and $y=r\cdot sin\theta$. Solve for $r$ $$\left\{ r=-\frac{b}{m\;\operatorname{cos}\left(\theta\right)-\operatorname{sin}\left(\theta\right)}\right\}$$

I typically use $0\le\theta\le 2\pi$ The idea of graphing a line in polar coordinates is more of a curiosity to me than anything practical. In Geogebra, (for example) there is not really a polar graphing system, but we can plot a polar equation by conversion to cartesian. that is, $x=r(\theta)\cos\theta$ and $y=r(\theta)\sin\theta$. Geogebra then superimposes a polar grid and viola it looks like we have graphed a polar equation. So... No real restriction on $r$, except that $\theta$ will have to have exceedingly high precision at certain points in its range or you will jump over the large values of $r(\theta)$. I would have to play with this awhile to give you any better answer.

Answer 3

The straight line through $(0,2)\text{ and } (3,0)$ has equation $$y=-\frac{2}{3}x+2$$ Since the the polar coordinates $(r,\alpha)$ correspond to $(r\sin\alpha,r\cos\alpha)$ this translates to $$ -\frac{2}{3}r\cos\alpha = r\sin\alpha+2$$

It only remains to solve for $r$.

Answer 4

Note also that (following from Niel's answer's that $\frac{x}{a}+\frac{y}{b}=1\iff r=\frac{a*b}{b*\cos(θ)+a*\sin(θ)}$) the Cartesian equation for the line joining two points $((x_0,y_0),(x_1,y_1))$ is $a=\frac{x_1*y_0-x_0*y_1}{y_0-y_1}, b=\frac{y_1*x_0-y_0*x_1}{x_0-x_1}$, so the polar equation is $\frac{\frac{x_1*y_0-x_0*y_1}{y_0-y_1}*\frac{y_1*x_0-y_0*x_1}{x_0-x_1}}{\frac{y_1*x_0-y_0*x_1}{x_0-x_1}*\cos(θ)+\frac{x_1*y_0-x_0*y_1}{y_0-y_1}*\sin(θ)}=\frac{x_0*y_1-x_1*y_0}{(y_1-y_0)*\cos(θ)+(x_0-x_1)*\sin(θ)}$.

Answer 5

Given the linear equation in two intercept form, $\frac{x}{a}+\frac{y}{b}=1$.

The equivalent polar form is $r=\frac{a*b*\sec(θ)}{b+\tan(θ)}$. Note, however, that $r=\frac{a*b}{b*\cos(θ)+a*\sin(θ)}$ requires one more variable lookup but is well-defined when $θ=π*n, n\in\mathbb{Z}$, instead of involving a division by $0$.

Answer 6

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

$$ r = p \sec \left( \theta - \tan^{-1} \frac{a}{b} \right ) $$