Convierte la ecuación polar a coordenadas rectangulares

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Hola Yani Juarez!

Equivalencias:

$$\begin{align}&r= \frac 1 {sen \theta-cox \theta}\\&\\&equivalencias:\\&r^2=x^2+y^2\\&\\&tan \theta= \frac y x ==> 1+tan^2 \theta=sec^2 \theta\\&\\&sec^2 \theta=1+ \frac {y^2}{x^2}=\frac {x^2+y^2}{x^2}==>\\&\cos^2 \theta= \frac {x^2}{x^2+y^2}\\&\cos \theta= \frac x { \sqrt {x^2+y^2}}\\&\\&sen \theta=tan \theta·\cos \theta= \frac y x· \frac x { \sqrt {x^2+y^2}}=\frac y { \sqrt {x^2+y^2}}\\&\\&r= \frac 1 {sen \theta-cox \theta}==> r(sen  \theta - \cos \theta)=1\\&\\&\sqrt{x^2+y^2}\Bigg(\frac y { \sqrt {x^2+y^2}}-\frac x { \sqrt {x^2+y^2}}\Bigg)=1\\&\\&y-x=1\\&\\&y=1+x\\&\\&\end{align}$$

Es una recta

Saludos

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