Question

Write the equation -2x+6y=7 in polar form.

Answer 1

What are the conversion relationships, relating x to radius r and angle theta, and relating y to radius r and angle theta?

Answer 2

x=?
y=? Write in the proper formulas.

Yes, you're right on that x.

Answer 3

x=r cos theta
y=r sin theta

Answer 4

Which I then got
-2 r cos theta+6 r sin theta=7, I just wasn't sure how to continue on

Answer 5

you could solve for r

Answer 6

but you have write it already in polar form

Answer 7

but if you wanted to solve for r
you could factor out the r from the two terms on the left hand side
and then divide by what is being multiplied by r on both sides

Answer 8

Would r(cos theta+sin theta)=-7/12 be correct?

Answer 9

what happen to -2 next to cos(theta) and the 6 next to the sin(theta)
also where did -1/12 come from

Answer 10

Wait...

Answer 11

Sorry, I was confused by what you were having me do, but I think I understand now..

Answer 12

\[r=\frac{7}{(-2cos\theta+6sin\theta)}\]

Answer 13

looks great

Answer 14

What would I do next?

Answer 15

nothing really you can do
other things you can do are just up to the viewer
like some math people don't like the term with - sign first
they might prefer to write this...since it cost less amount of symbols:
\[r=\frac{7}{6 \sin(\theta)-2 \cos(\theta)}\]

Answer 16

Well, my question is having me write it like http://imgur.com/Xd97J2h

Answer 17

\[\sqrt{6^2+(-2)^2}=\sqrt{6^2+4^2}=\sqrt{36+16}=\sqrt{52}=\sqrt{4 \cdot 13}=2 \sqrt{13} \\ \text{ so anyways} \\ 2 \sqrt{13}(\frac{6}{2 \sqrt{13}} \sin(\theta)-\frac{2 }{2 \sqrt{13}} \cos(\theta)) =... \text{ one sec }...\]



so we have
\[2 \sqrt{13}(\frac{6}{2 \sqrt{13}} \sin(\theta)+\frac{-2}{2 \sqrt{13}} \cos(\theta)) \\ =2 \sqrt{13}(\cos(u) \sin(\theta)+\sin(u)\cos(\theta)) \\ = 2 \sqrt{13}( \text{ hint: use a sum identity here })\]