Convert \(r = {7 \over 9sinθ-cosθ}\) to rectangular form.

Enter your answer in slope-intercept form by filling in the boxes. Enter values so that fractions are simplified.

\(y = { [] \over []} x + \frac{[]}{[]}\)



Hello:) I'm trying to convert the polar equation above into a rectangular equation. I looked up a Youtube video to do it and I got stuck after a certain point.



Here's my work:


1) I multiplied r on both sides.

\(r^2 = {7 \over 9sinθ - cosθ} \times r\)


2) Because r2 equals x2 + y2 , I substituted x2 + y2​ into r2.

\(x^2 + y^2= {7 \over 9sinθ-cosθ}\times r\)



For the next part, I'm confused, because it says that cosθr is equal to x. And because the r is being multiplied at the end, I'm tripped up lol so if someone could help me understand the next step, I'd appreciate that so much!

 Jun 4, 2020

I'm not sure about the approach you took, but have you tried multiplying both sides by the denominator(\(9\sin{\theta}-\cos{\theta}\)) from the beginning? Maybe that would get you somewhere, since you know that \(r * \cos{\theta} = x \) and \(r * \sin{\theta} = y\)


Not sure if you're asking for the solution, but if you are, then you can keep reading I guess.


Using the method mentioned above:


\(9r\sin{\theta} - r\cos{\theta} = 7\)

Using our substitutions, we rewrite this as:

\(9y - x = 7\)

From here on out, it gets pretty intuitive. Dividing and rearranging to get our desired form, we get:

\(9y = x+7\)

\(y = x/9 + 7/9\)


\(y = \frac19 x + \frac79\)


Hope this helped!

 Jun 4, 2020
edited by jfan17  Jun 4, 2020

Omigosh! I can see where I went wrong. I should've seen that lol. I understand 100% now. Thank you so so so much. You're a life savior!! Truly:)

auxiarc  Jun 4, 2020

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