How do you convert $r = 6 / (2cos[theta] - 3sin[theta])$ into cartesian form?

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Answer 1 · 2018-04-28T11:55:14

$2x - 3y=6$

We know our polar conversions:

$r^2 = x^2 + y^2$

$rcostheta = x$

$rsintheta = y$

$r = 6/ (2costheta - 3sintheta)$

$=> 2rcostheta - 3rsintheta = 6$

$=>color(red)( 2x - 3y = 6$

How do you convert r = 6 / (2*cos[theta] - 3*sin[theta]) r = 6 / (2*cos[theta] - 3*sin[theta]) into cartesian form?