How do you write $3 x - 4 y - 10 = 0$ $3x-4y-10=0$ in polar form?

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Answer 1 · 2017-11-06T17:33:26

$r = \frac{10}{3 cos θ - 4 sin θ}$ $r=10/(3costheta-4sintheta)$

we have the conversion eqns

$r^{2} = x^{2} + y^{2} - - (1)$ $r^2=x^2+y^2--(1)$

$x = r cos θ - - (2)$ $x=rcostheta--(2)$

$y = r sin θ - - - (3)$ $y=rsintheta---(3)$

we have

$3 x - 4 y - 10 = 0$ $3x-4y-10=0$

using $(2) & (3)$ $(2)" & "(3)$

$3 r cos θ - 4 r sin θ - 10 = 0$ $3rcostheta-4rsintheta-10=0$

$r (3 cos θ - 4 sin θ) = 10$ $r(3costheta-4sintheta)=10$

$r = \frac{10}{3 cos θ - 4 sin θ}$ $r=10/(3costheta-4sintheta)$

How do you write 3x-4y-10=03x−4y−10=03x-4y-10=0 in polar form?