How do you convert #(-5,0) # to polar form?

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Answer 1 · 2017-12-15T14:12:02

#(5,pi)#

Using:

#x=rcostheta#

#y=rsintheta#

#theta=arctan(y/x)#

From example:

#-5=rcosthetacolor(white)(88)[1]#

#0=rsinthetacolor(white)(8888)[2]#

Squaring [1] and [2]:

#25=r^2cos^2theta#

#0=r^2sin^2theta#

Adding [1] and [2]:

#25=r^2cos^2theta+r^2sin^2theta#

Factor:

#25=r^2(cos^2theta+sin^2theta)#

#25=r^2=> r=+-5# ( use #r=5# )

For #r=5#:

#y=0=5sintheta=>sintheta=0#

#x=-5=5costheta=>costheta=-1#

#theta=arctan(sintheta/costheta)=0/-1= , pi#

#(5,pi)#