Let's solve the problem step by step. ### **Given:** - Cartesian (rectangular) coordinates: \( (6, -2) \) - Find **two sets** of polar coordinates \((r, \theta)\) for the same point, where \( 0 \leq \theta < 2\pi \). --- ## **Step 1: Convert (6, -2) to Polar Coordinates** The formulas for converting are: - \( r = \sqrt{x^2 + y^2} \) - \( \theta = \arctan\left(\frac{y}{x}\right) \) Given \( x = 6, y = -2 \): ### a. Calculate \( r \): \[ r = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \] ### b. Find \( \theta \): \[ \theta = \arctan\left(\frac{-2}{6}\right) = \arctan\left(-\frac{1}{3}\right) \] This angle is in the fourth quadrant because \( x > 0 \) and \( y < 0 \). \[ \arctan\left(-\frac{1}{3}\right) \approx -0.3218 \text{ radians} \] But to get the value in \( [0, 2\pi) \) range: \[ \theta = 2\pi + (-0.3218) \approx 6.2832 - 0.3218 = 5.9614 \text{ radians} \] --- ## **Step 2: Write the First Set of Polar Coordinates** \[ \boxed{\left(2\sqrt{10},\, 5.961 \right)} \] or, if you want to write the angle as a negative: \[ \left(2\sqrt{10},\, -0.322\right) \] But for \( 0 \leq \theta < 2\pi \), use \( 5.961 \) radians. --- ## **Step 3: Find Another Polar Representation** Recall: - The point \((r, \theta)\) is the same as \((-r, \theta + \pi)\). So, take \( r = -2\sqrt{10} \), and add \( \pi \) to \( \theta \): \[ \theta_2 = 5.961 + \pi = 5.961 + 3.142 = 9.103 \] But since \( \theta_2 \) must be in \( [0, 2\pi) \): \[ 9.103 - 2\pi = 9.103 - 6.283 = 2.82 \text{ radians} \] So the second set is: \[ \boxed{\left(-2\sqrt{10},\, 2.82 \right)} \] --- ## **Final Answers** ### **Two sets of polar coordinates for the point \( (6, -2) \) are:** \[ \boxed{ \begin{aligned} &1.\quad \left(2\sqrt{10},\, 5.96\right) \\ &2.\quad \left(-2\sqrt{10},\, 2.82\right) \end{aligned} } \] Where angles are in radians (rounded to two decimal places).

Let's go through the problem in detailed steps, including all calculations and explanations. --- ## **Given:** - Cartesian point: \( (x, y) = (6, -2) \) - Goal: 1. Plot the point on the rectangular plane. 2. Find **two sets** of polar coordinates \((r, \theta)\) that correspond to this same point, with \( 0 \leq \theta < 2\pi \). --- ## **Step 1: Understand the conversion from Cartesian to Polar coordinates** The polar coordinates \((r, \theta)\) are related to Cartesian coordinates via: \[ r = \sqrt{x^2 + y^2} \] \[ \theta = \arctan\left(\frac{y}{x}\right) \] However, \( \arctan \) alone gives an angle in the range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). We need to consider the signs of \(x\) and \(y\) to determine the correct quadrant. --- ## **Step 2: Calculate \( r \)** \[ r = \sqrt{x^2 + y^2} = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} \] Simplify: \[ r = \sqrt{40} = \sqrt{4 \times 10} = 2 \sqrt{10} \] **Numerical value:** \[ r \approx 2 \times 3.1623 = 6.3246 \] --- ## **Step 3: Calculate \( \theta \)** \[ \theta' = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{-2}{6}\right) = \arctan\left(-\frac{1}{3}\right) \] Calculate: \[ \arctan\left(-\frac{1}{3}\right) \approx -0.3218 \text{ radians} \] **Note:** This is the principal value, which is negative and in the 4th quadrant. Since: - \( x = 6 > 0 \) - \( y = -2 < 0 \) The point is located in the **4th quadrant**. To express \( \theta \) in the range \( [0, 2\pi) \): \[ \theta = 2\pi + \theta' \approx 6.2832 - 0.3218 = 5.9614 \text{ radians} \] --- ## **Step 4: First polar coordinate set** \[ \boxed{ \left(r, \theta\right) = \left(2 \sqrt{10},\, 5.9614 \text{ radians}\right) } \] or approximately: \[ \boxed{ \left(6.3246,\, 5.96\right) } \] --- ## **Step 5: Find the second set of coordinates** Recall that: \[ (r, \theta) \quad \text{and} \quad (-r, \theta + \pi) \] represent the same point in the plane, provided the angle is adjusted to stay within \( [0, 2\pi) \). - Starting with \( r = 2 \sqrt{10} \), \( \theta \approx 5.9614 \). Now, consider: \[ r' = -r = -2 \sqrt{10} \] \[ \theta' = \theta + \pi \] Calculate: \[ \theta' = 5.9614 + 3.1416 \approx 9.103 \text{ radians} \] But since \( \theta' \) must be in \( [0, 2\pi) \), subtract \( 2\pi \): \[ 2\pi \approx 6.2832 \] \[ \theta'_{\text{mod}} = 9.103 - 6.2832 \approx 2.82 \text{ radians} \] Hence, the second set of coordinates: \[ \boxed{ \left(-2 \sqrt{10},\, 2.82 \text{ radians}\right) } \] --- ## **Final Answers:** ### **First set:** \[ \boxed{ \left( 2\sqrt{10},\, 5.96 \text{ radians} \right) } \] ### **Second set:** \[ \boxed{ \left( -2\sqrt{10},\, 2.82 \text{ radians} \right) } \] **Note:** The angles are rounded to two decimal places for clarity. --- ## **Summary:** - The point \( (6, -2) \) in Cartesian coordinates corresponds to: - \( \left( 2\sqrt{10},\, 5.96 \right) \) - \( \left( -2\sqrt{10},\, 2.82 \right) \) - These two sets of polar coordinates represent the same point, with the second including a negative radius and an adjusted angle. --- **Let me know if you'd like any further clarification!**