# The Cartesian coordinates of a point are given:

# (i) Find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π.

# (ii) Find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (2, -2).

The system which we have described to label points in a plane is known as the Cartesian System.

## Answer: i) Polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π are (2√2, 7π / 4), ii) Polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π (2, -2) are (-2√2, 3π / 4)

We will make use of the concept of the Pythagoras theorem to find the polar coordinates.

**Explanation:**

i) Apply the Pythagoras theorem : r² = x² + y²

r = √2² + (- 2)² = √4 + 4 = √8 = 2√2 [ substituting x = 2 and y = -2 in the given pythagoras theorem ]

Since r > 0, we can write r = 2√2

Recall that:

r cos θ = x, r sin θ = y

Therefore, 2√2 cos θ = 2, 2√2 sin θ = -2

Divide both sides by 2√2

cos θ = 1/√2 → Equation 1

sin θ = - 1/√2 → Equation 2

Now, Equation 1 and 2 are satisfied by θ = 7π / 4

Polar Coordinates of the points are (2√2, 7π / 4)

The point (2,-2) is in the 4th quadrant. In the 4th quadrant tan is -1 at θ = 7π / 4 when 0 ≤ θ < 2π

ii) Apply the Pythagoras theorem r² = x² + y²

r = √2² + ( -2 )² = √4 + 4 = √8 = 2√2

Since r < 0, we can write r = -2√2

Recall that:

r cos θ = x, r sin θ = y

Therefore, -2√2 cos θ = 2, -2√2 sin θ = -2

Divide both sides by -2√2

cos θ = -1/√2 → Equation 1

sin θ = 1/√2 → Equation 2

Now, Equation 1 and 2 are satisfied by θ = 3π / 4

Polar Coordinates of the points are ( -2√2, 3π / 4 )

In the 2nd quadrant tan θ is -1 at θ = 3π / 4 when 0 ≤ θ < 2π.

### Thus, the Polar Coordinates of the points are (2√2, 7π/4), (-2√2, 3π / 4 ).

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