What is the polar form of #(-3,4)#?

1 Answer
Jan 6, 2016

#(5,126.87^0)#

Explanation:

If #(a,b)# is a are the coordinates of a point in Cartesian Plane, #u# is its magnitude and #alpha# is its angle then #(a,b)# in Polar Form is written as #(u,alpha)#.
Magnitude of a cartesian coordinates #(a,b)# is given by#sqrt(a^2+b^2)# and its angle is given by #tan^-1(b/a)#

Let #r# be the magnitude of #(-3,4)# and #theta# be its angle.
Magnitude of #(-3,4)=sqrt((-3)^2+4^2)=sqrt(9+16)=sqrt25=5=r#
Angle of #(-3,4)=Tan^-1(4/-3)=Tan^-1(-4/3)=-53.13^0#
#implies# Angle of #(-3,4)=-53.13^0#

But the given point (-3,4) is in second quadrant so we have to add #180^0# in the angle.
#implies# Angle of #(-3,4)=-53.13^0+180^0=126.87^0=theta#

#implies (-3,4)=(r,theta)=(5,126.87^0)#
#implies (-3,4)=(5,126.87^0)#
Note that the angle is given in degree measure.