How do you convert #y = -x# to polar form?

1 Answer
May 3, 2016

Half-line #theta=(3pi)/4#, pole r=0 and the opposite half-line #theta=(7pi)/4#.

Explanation:

Cartesian #( x, y ) = ( r cos theta, r sin theta )# is #( r, theta )#, in polar form.

So, the equation #y = - x# becomes

#r sin theta=r cos theta#.

The solutions are # r = 0 and sin theta = - cos theta#

Further, in [0, #2pi#], #theta=(3pi)/4 and theta=(7pi)/4# make #sin theta = - cos theta#.

Importantly, r = 0 should be included, as .#theta=(3pi)/4 and theta=(7pi)/4# represent radial lines, sans r= 0.

In polar coordinates, r = 0 represents a null vector that has the distinction of associating with any direction..In the absence of r = 0, the line has discontinuity at the pole.

So, unless necessitated, polar form of the equation of a straight line is to be avoided. .