What is the polar form of #( -7,-1 )#?

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Answer 1 · 2017-07-06T00:41:09

#(sqrt50,3.28)#

To convert this to a polar coordinate #(r, theta)#, you can use the following formulas and substitute #-7# for #x# and #-1# for #y#.

#r^2 = x^2 + y^2#
#tan theta = (y)/(x)#

#r^2 = (-7)^2 + (-1)^2#
#r^2 = 49 + 1#
#r^2 = 50#
#r = sqrt50#

#tan theta = (y)/(x)#
#tan theta = (-1)/(-7)#
#theta = tan^-1(1/7)#
#theta ~~ 0.14#

Since the coordinate is in quadrant #"III"#, we must add #pi# to this for the correct angle:

#= 0.14 + pi ~~ 3.28#

Thus, the polar form of #(-7,-1)# is #(sqrt50,3.28)#.

Answer 2 · 2017-07-06T00:48:55

#(sqrt(50), 3.283)# (in radians)

The polar form of a rectangular coordinate is given by

#r = sqrt(x^2 + y^2)#

#theta = arctan(y/x)#

So,

#r = sqrt((-7)^2 + (-1)^2) = color(red)(sqrt(50)#

#theta = arctan((-1)/(-7)) = 0.142# #"rad"# #+ pi = color(blue)(3.283# #color(blue)("rad"#

#pi# was added because the coordinate is in the third quadrant.

The polar form is thus

#(color(red)(sqrt(50)), color(blue)(3.283))#