How do you convert $x + y = 9$ $x+y=9$ to polar form?

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Answer 1 · 2016-12-31T09:28:21

$r = \frac{9}{\sqrt{2}} sec (θ - \frac{π}{4})$ $r=9/sqrt2 sec(theta-pi/4)$

If $N (p, α)$ $N(p, alpha)$ is the foot of the perpendicular to the straight line

from the pole r = 0 and $P (r, θ)$ $P(r, theta)$ on the line, the equation to the

line is

$r = p sec (θ - α)$ $r=psec(theta-alpha)$ ,

using the projection $O P cos ∠ P O N = O N$ $OPcos anglePON=ON$ .

In cartesian form, this is

$x cos α + y sin α = p$ $xcos alpha + y sin alpha = p$

Here, $p = \frac{9}{\sqrt{2}} and α = \frac{π}{4}$ $p = 9/sqrt2 and alpha = pi/4$ , and so, the equations are

$r = \frac{9}{\sqrt{2}} sec (θ - \frac{π}{4})$ $r = 9/sqrt2 sec(theta-pi/4)$ and

$\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = \frac{9}{\sqrt{2}}$ $x/sqrt2+y/sqrt2=9/sqrt2$ .

However, directly ( ignoring all these relevant details, about the

polar form )

$x + y = r (cos θ + sin θ) = 9$ $x+ y = r(cos theta+sin theta)=9$ . Simplifying for explicit r

$r = \frac{9}{\sqrt{2}} sec (θ - \frac{π}{4}) = \frac{9}{\sqrt{2}} csc (θ + \frac{π}{4})$ $r =9/sqrt2sec(theta-pi/4)=9/sqrt2csc(theta+pi/4)$

How do you convert x+y=9x+y=9 x+y=9 to polar form?