To find the polar coordinates of a point given in Cartesian coordinates $(x, y)$, the formulas are:

1. $r = \sqrt{x^2 + y^2}$ (distance from the origin)
2. $\theta = \tan^{-1}\left(\frac{y}{x}\right)$ (angle in radians or degrees)

Here, the given Cartesian coordinates are $(3x, -3y)$. Using the formulas:

Step 1: Find $r$

$r = \sqrt{(3x)^2 + (-3y)^2} = \sqrt{9x^2 + 9y^2} = \sqrt{9(x^2 + y^2)} = 3\sqrt{x^2 + y^2}$

Since $r$ is provided as $14.4$, we know: $3\sqrt{x^2 + y^2} = 14.4 \implies \sqrt{x^2 + y^2} = 4.8$

Step 2: Find $\theta$

$\theta = \tan^{-1}\left(\frac{-3y}{3x}\right) = \tan^{-1}\left(\frac{-y}{x}\right)$

The quadrant will determine the angle:

• Since $3x > 0$ and $-3y < 0$, the point lies in Quadrant IV, where the angle is measured as: $\theta = 360^\circ - |\tan^{-1}(\frac{y}{x})|$

Let me know the specific values of $x$ and $y$ if you need further numerical calculations or have additional data!

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Questions to Expand Learning:

1. What is the significance of polar coordinates in real-world applications?
2. Can Cartesian coordinates always be converted into polar coordinates?
3. How does the quadrant of a point affect the angle $\theta$ in polar coordinates?
4. What are the potential uses of polar coordinates in engineering or physics?
5. How do you interpret negative angles in polar coordinates?

Tip:

When converting Cartesian to polar coordinates, always check the quadrant of the point to adjust the angle $\theta$ appropriately.