How do you convert $r = \frac{6}{2 \cdot cos [θ] - 3 \cdot sin [θ]}$ $r = 6 / (2cos[theta] - 3sin[theta])$ into cartesian form?

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Answer 1 · 2018-04-28T11:55:14

$2 x - 3 y = 6$ $2x - 3y=6$

We know our polar conversions:

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

$r cos θ = x$ $rcostheta = x$

$r sin θ = y$ $rsintheta = y$

$r = \frac{6}{2 cos θ - 3 sin θ}$ $r = 6/ (2costheta - 3sintheta)$

$\Rightarrow 2 r cos θ - 3 r sin θ = 6$ $=> 2rcostheta - 3rsintheta = 6$

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

How do you convert r = 6 / (2*cos[theta] - 3*sin[theta]) r=62⋅cos[θ]−3⋅sin[θ]r = 6 / (2*cos[theta] - 3*sin[theta]) into cartesian form?