The cartesian coordinates of a point are given. (a) (2, −2) (i) find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ) = (ii) find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ) = (b) (−1, 3 ) (i) find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ) = (ii) find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ) =

We know that
the relationship between the 2-dimensional polar and Cartesian coordinates is

r = √(x² + y²)

Θ = tan⁻¹ (y/x)
so

Part a) (2, −2)> this point belong to the IV quadrant
r = √(x² + y²) r = √(2² + (-2)²)> r=√8
Θ = tan⁻¹ (y/x) Θ = tan⁻¹ (2/2)> 45°
remember that the point belong to the IV quadrant
so
Θ=360-45> Θ=315°

the answer part A) is 
(r,Θ)=(√8,315°)

Part b) (-1, 3)> this point belong to the II quadrant
r = √(x² + y²) r = √(-1² + (3)²)> r=√10
Θ = tan⁻¹ (y/x) Θ = tan⁻¹ (3/1)> 71.57°
remember that the point belong to the II quadrant
so
Θ=180-71.57> Θ=108.43°

the answer part B) is 
(r,Θ)=(√10,108.43°)

127 millimeters is the answer