How do you convert #(-sqrt6, -sqrt2)# to polar form?

1 Answer
KillerBunny
Oct 6, 2016

#(-sqrt6,-sqrt2)# in cartesian coordinate is #(sqrt(8),7/6pi)# in polar coordinates

Explanation:

You have a point expressed in cartesian coordinates #(x,y)#, and you need its polar coordinates.

Polar coordinates find points by knowing the length of the line that connects the point to the origin, and the angle that line forms with the #x# axis.

The length of the line that connects #(0,0)# to #(-sqrt6, -sqrt2)# is given by Pytagora's theorem:

#rho = sqrt( (0-sqrt6)^2+ (0-sqrt2)^2 ) = sqrt(6+2)=sqrt(8)=2sqrt(2)#

As for the angle, the formula states that

#theta = arctan(y/x) = arctan((-sqrt2)/-sqrt6) = arctan(sqrt(1/3))=pi/6#

Actually, this formula works in the first quadrant. Since we are in the third one, we need to add #180# degree to our angle, because #x# and #x+pi# have the same tangent value. So, our true angle will be #pi/6+pi = 7/6pi#

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