How do you convert #(-2, 5)# to polar form?

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Answer 1 · 2017-12-10T13:52:14

#(sqrt(29), pi-tan^(-1)(5/2))# or #(-sqrt(29), tan^(-1)(-5/2))#

To convert rectangular #(a,b)# to polar we use the two formulas:

#r=sqrt(a^2+b^2)# and #hat theta=tan^(1)(|b/a|)#

then we have to "fix" the quadrant for #theta#.

#r=sqrt((-2)^2+5^2)=sqrt(29)#

#hat theta = tan^(-1)(5/2)#

the point #(-2,5)# is in QII, so #theta = pi-hat theta#

therefore #theta = pi-tan^(-1)(5/2)#.

An alternate, but equivalent representation is #(-sqrt(29), tan^(-1)(-5/2))#, which uses a QIV angle but a negative #r# to end up in QII.