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.
