Hello, stryker213!

We are given: .x = -3, .y = 3√3. .The point is in Quadrant 2.1. For a given rectangle coordinates, find two pairs of polar coordinates for the point.

One pair satisfies r>0 and 0 ≤ θ < 2π,

and the second pair satisfies r ≥ 0 and -2π < θ ≤ 0

for the problem (-3, 3√3)

. . . . . . . . . .____________

Then: .r .= .√(-3)² + (3√3)² .= .6

And: .tanθ .= .(3√3)/(-3) .= .-√3 . → . θ = 2π/3

The polar coordinates are: .(6, 2π/3) and (6, -4π/3)

You're expected to know the conversion formulas: .x .= .r·cosθ, .y .= .r·sinθ2. Identify the curve by transforming r·sinθ = 4 into rectangular coordinates.

Then .r·sinθ = 4 .becomes .y = 4, a horizontal line.