In order to transform this straight line equation:
y = x + 5,
to its polar form with the radius vector 'r' in function of the vectorial theta angle. In order to transform this straight line equation:
y = x + 5,
to its polar form with the radius vector 'r' in function of the vectorial theta angle. @Mathematics

Question

chriss · Answer

I am not exactly sure what you are asking, but

$$r \sin (\theta)=r \cos (\theta)+5$$

chriss · Answer

$$y=r \sin (\theta)$$
and
$$x=r \cos (\theta)$$

chriss · Answer

I guess you are trying to find $$(r,\theta)$$

liliakarina · Answer

That's exactly what I was trying to figure out. Sorry 'cause I did not formulate correctly the answer. Thanks.

chriss · Answer

I'm working through it.  I have r.  It is 
$$r=\frac{5}{\sin(\theta)-\cos(\theta)}$$

I arrived at it like this:

$$rsin(\theta)=rcos(\theta)+5$$
$$rsin(\theta)-rcos(\theta)=5$$
$$r(\sin(\theta)-\cos(\theta))=5$$
$$r=\frac{5}{\sin(\theta)-\cos(\theta)}$$

Give me a moment on theta.

liliakarina · Answer

Cool, okay.

chriss · Answer

I'm a little stuck here.  I'm not really sure how to isolate theta in this equation.  I'll keep at it, but I don't have a solution for theta at the moment.

liliakarina · Answer

It's okay, thank you.

chriss · Answer

you're welcome... at the moment I have it down to 
$$\sin(\theta)-\cos(\theta)=\frac{5}{r}$$

I'm just not sure where to go with it from there.

liliakarina · Answer

All right, I'll figure out from here.