How do you write the equation $3 x + 2 y - 5 = 0$ $3x+2y-5=0$ in polar form?

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Answer 1 · 2017-05-06T12:30:33

$r = \frac{5}{3 cos θ + 2 sin θ}$ $r=5/(3costheta+2sintheta)$

for polar coordinates we use the relations

#x^2+y^2=r^2

$x = r cos θ$ $x=rcostheta$

$y = r sin θ$ $y=rsintheta$

$3 x + 2 y - 5 = 0$ $3x+2y-5=0$

becomes

$3 r cos θ + 2 r sin θ - 5 = 0$ $3rcostheta+2rsintheta-5=0$

$r (3 cos θ + 2 sin θ) = 5$ $r(3costheta+2sintheta)=5$

$r = \frac{5}{3 cos θ + 2 sin θ}$ $r=5/(3costheta+2sintheta)$

How do you write the equation 3x+2y-5=03x+2y−5=03x+2y-5=0 in polar form?