How do you convert the polar equations in Cartesian form #Theta=-pi/6#?

Question

2451 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-19T14:36:49

4th quadrant half-line of #y = -x/sqrt 3#.

The line #theta=-pi/6# is the radial line in the 4th quadrant, wherein

the polar coordinate# r >=0# is arbitrary.

Use #(x, y)=r(cos theta, sin theta)#.

Here, #x =r cos(-pi/6)=r sqrt 3/2 and y=r sin (-pi/6)=-r/2, r>=0#

In brief, #(x, y)=r(sqrt 3/2, -1/2)#.

and this represents the 4th quadrant half-line of the whole line

#y = -x/sqrt 3#.

The other half line in the 2nd quadrant is given by #theta=(5pi)/6#

and the separate cartesian (parametric) equations are

#(x, y)= r(sqrt 3/2, -1/2)# and

.#(x, y)= r(-1/2, sqrt 3/2)#, r>=0..