What is the polar form of #( -5,-1 )#?

1 Answer
Jan 3, 2016

#(sqrt26,arctan(1/5) - pi)#

Explanation:

Let #A(-5,-1)#.The polar form will be something like #(r,theta)# with r non-negative and #theta in [0,2pi]#.

The module will be given by the norm of the vector #OA# which is #sqrt((-5)^2 + (-1)^2) = sqrt26#.

The angle between the #(Ox)# axis and the vector #OA# will be given by #arctan(y/x) - pi = arctan((-1)/(-5)) - pi = arctan(1/5) - pi# (we substract #pi# because #x < 0# and #y < 0#, and it will give us the principal measure of the angle ie the angle in #]-pi,pi]#).