How do you convert $x + 2 y - 4 = 0$ $x+2y-4=0$ to polar form?

Question

3503 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-04T16:15:32

$r = 4 \frac{cos α}{cos (θ - α)}$ $r=4cosalpha/cos(theta-alpha)$ where $α = {tan}^{- 1} 2$ $alpha=tan^(-1)2$

If $(r, θ)$ $(r,theta)$ is in polar form and $(x, y)$ $(x,y)$ in Cartesian form the relation between them is as follows:

$x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ , $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$ and $tan θ = \frac{y}{x}$ $tantheta=y/x$

Hence, $x + 2 y - 4 = 0$ $x+2y-4=0$ can be written as

$r cos θ + 2 r sin θ - 4 = 0$ $rcostheta+2rsintheta-4=0$ or

$r (cos θ + 2 sin θ) = 4$ $r(costheta+2sintheta)=4$ or

$r = \frac{4}{cos θ + 2 sin θ}$ $r=4/(costheta+2sintheta)$ .....(A)

Now let ${tan}^{- 1} 2 = α$ $tan^(-1)2=alpha$ or $2 = tan α = \frac{sin α}{cos α}$ $2=tanalpha=sinalpha/cosalpha$

Hence (A) becomes $r = \frac{4}{cos θ + (\frac{sin α}{cos α}) sin θ}$ $r=4/(costheta+(sinalpha/cosalpha)sintheta)$

or $r = 4 \frac{cos α}{cos θ cos α + sin α sin θ}$ $r=4cosalpha/(costhetacosalpha+sinalphasintheta)$ or

$r = 4 \frac{cos α}{cos (θ - α)}$ $r=4cosalpha/cos(theta-alpha)$

How do you convert x+2y-4=0x+2y−4=0x+2y-4=0 to polar form?