Question

Given point m(6,-6 sqrt(3)) which is located on the rectangular coordinate plane, what are the representations of the point in polar form where -2 pi ≤θ ≤2 pi?

Answer 1

To represent the point (6, -6 sqrt(3)) in polar form, find the distance from the origin and the angle it makes with the positive x-axis. The point in polar form is (12, arctan(-sqrt(3))).

Step-by-step explanation:

To represent the point (6, -6 sqrt(3)) in polar form, we need to find the distance from the origin and the angle it makes with the positive x-axis. First, let's find the distance using the distance formula:

d = sqrt((6)^2 + (-6 sqrt(3))^2)

Simplifying, we get d = sqrt(36 + 108) = sqrt(144) = 12.

Now, let's find the angle. We can use the inverse tangent function:

theta = arctan((-6 sqrt(3))/6)

Simplifying, we get theta = arctan(-sqrt(3)).

Therefore, the point (6, -6 sqrt(3)) in polar form is (12, arctan(-sqrt(3))).