Find the polar coordinates, \( -\pi \leq \theta<\pi \) and \( r \leq 0 \), of the following points given in Cartesian coordinates. \( \begin{array}{llll}\text { a. }(-3,0) & \text { b. }(3,0) & \text { c. }(0,-1) & \text { d. }\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) \\ \text { a. The point }(-3,0) \text { in Cartesian coordinates is } \square \text { in polar coordinates for }-\pi \leq \theta<\pi \text { and } r \leq 0\end{array} \) (Type an ordered pair. Type an exact answer, using \( \pi \) as needed.)

To convert Cartesian coordinates to polar coordinates, we use the formulas: \[ r = \sqrt{x^2 + y^2} \] \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \] However, we need to adjust \( r \) and \( \theta \) to meet the conditions \( r \leq 0 \) and \( -\pi \leq \theta < \pi \). Let's find the polar coordinates for each point: ### a. \((-3, 0)\) 1. Calculate \( r \): \[ r = \sqrt{(-3)^2 + 0^2} = \sqrt{9} = 3 \] 2. Calculate \( \theta \): \[ \theta = \tan^{-1}\left(\frac{0}{-3}\right) = \pi \quad (\text{since the point is on the negative x-axis}) \] Since we need \( r \leq 0 \), we take \( r = -3 \) and adjust \( \theta \): \[ \theta = \pi \] Thus, the polar coordinates for \((-3, 0)\) are: \[ \boxed{(-3, \pi)} \] ### b. \((3, 0)\) 1. Calculate \( r \): \[ r = \sqrt{3^2 + 0^2} = \sqrt{9} = 3 \] 2. Calculate \( \theta \): \[ \theta = \tan^{-1}\left(\frac{0}{3}\right) = 0 \] Since we need \( r \leq 0 \), we take \( r = -3 \) and adjust \( \theta \): \[ \theta = \pi \] Thus, the polar coordinates for \((3, 0)\) are: \[ \boxed{(-3, \pi)} \] ### c. \((0, -1)\) 1. Calculate \( r \): \[ r = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1 \] 2. Calculate \( \theta \): \[ \theta = \tan^{-1}\left(\frac{-1}{0}\right) = -\frac{\pi}{2} \quad (\text{since the point is on the negative y-axis}) \] Since we need \( r \leq 0 \), we take \( r = -1 \) and keep \( \theta \) as: \[ \theta = -\frac{\pi}{2} \] Thus, the polar coordinates for \((0, -1)\) are: \[ \boxed{(-1, -\frac{\pi}{2})} \] ### d. \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\) 1. Calculate \( r \): \[ r = \sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{2}{4} + \frac{2}{4}} = \sqrt{1} = 1 \] 2. Calculate \( \theta \): \[ \theta = \tan^{-1}\left(\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\right) = \tan^{-1}(1) = \frac{\pi}{4} \] Since we need \( r \leq 0 \), we take \( r = -1 \) and adjust \( \theta \): \[ \theta = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \] Thus, the polar coordinates for \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\) are: \[ \boxed{(-1, \frac{5\pi}{4})} \]