Polar equations

I have the equation x - y = 3 and need to find its polar equivalent.
I got to the expression r = 3 / (cos theta - sin theta) but the mark scheme seems to divide r(cos theta - sin theta) = 3 by root 2 and then manipulate to get (3 over root 2) sec(theta plus pi over 4)

Would my answer still be correct, and how does their one work?

Reply 1

old_engineer

It looks as though the mark scheme has simply used cos@ - sin@ = Rcos(@ + a), and found the values of R and a.

Although your answer is correct, I don't know whether it would have gained full marks. The mark scheme answer is arguably more useful as it provides an easier route to finding the value(s) of @ corresponding to a given value of r.