Find the equation in polar coordinates of the line through the point (−2,−3) and with slope 1/7

Question

I'm having some trouble with my assignment problem. I have no idea where to start at the moment and would appreciate any tips to help me get started on solving this problem.

Question

I'm asked to find the equation in polar coordinates of the line through the point (-2,-3) and with slope 1/7

Answer 1

Hint: First, find the equation of the line in Cartesian ($y=mx+b$ (shouldn't be very difficult)). You know the slope is $\frac{1}{7}=m$. They gave you another point that it goes through/satisfies. Use that to find $b$ and you'll have the equation of your line in Cartesian Coordinates. Then, use the relationship between Cartesian and Polar to get the polar equation (maybe $x=r\cos(\theta)$ and $y=r\sin(\theta)$ will ring a bell? (try to just drawing a triangle and review what Polar coordinates and Cartesian coordinates mean)) Good luck.

Answer 2

Hint:

You can write the equation in cartesian coordinate as $y=mx+q$, than substitute $y=\rho \sin \theta$ and $x=\rho \sin \theta$. From this it is easy to obtain $\rho$ as a function of $\theta$.

Answer 3

we have the equation in cartesian coordinates as $$y=\frac{1}{7}x-\frac{19}{7}$$ now we have $$\cos(\theta)=\frac{x}{r}$$ and $$\sin(\theta)=\frac{y}{r}$$ can you finish?

Answer 4

hint

Your line has the following cartesian equation

$$y=\frac {1}{7}x+b $$

with $$b=-3+\frac {2}{7}=-\frac {19}{7} $$

thus

$$r=\sqrt {x^2+y^2}=f (x) $$

and $$\tan (\theta)=\frac {y}{x} =g (x)$$ or $$x=h (\theta) $$

thus, your polar equation will be $$r=f (h (\theta)). $$