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

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

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 r^{2} equals x^{2} + y^{2} , I substituted x^{2} + y^{2} into r^{2}.

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

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!

auxiarc Jun 4, 2020

#1**+1 **

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:

\(9y = x+7\)

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

or

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

Hope this helped!

.jfan17 Jun 4, 2020