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

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Answer 1 · 2016-12-31T09:28:21

#r=9/sqrt2 sec(theta-pi/4)#

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

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

line is

#r=psec(theta-alpha)#,

using the projection #OPcos anglePON=ON#.

In cartesian form, this is

#xcos alpha + y sin alpha = p#

Here, #p = 9/sqrt2 and alpha = pi/4#, and so, the equations are

#r = 9/sqrt2 sec(theta-pi/4)# and

#x/sqrt2+y/sqrt2=9/sqrt2#.

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

polar form )

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

#r =9/sqrt2sec(theta-pi/4)=9/sqrt2csc(theta+pi/4)#