Let's solve each problem one by one.

Problem 1:

In $\mathbb{R}^2$, find the equation of the line symmetric to $y = x + 2$ with respect to the line $y = 2x$.

To find the equation of the line symmetric to $y = x + 2$ with respect to $y = 2x$:

1. Convert $y = x + 2$ and $y = 2x$ into parametric forms.
2. Reflect each point on the line $y = x + 2$ over the line $y = 2x$.
3. Derive the new line's equation.

Solution:

The line $y = x + 2$ can be written parametrically as: $(x, x + 2)$

For reflection across $y = 2x$, we use the following transformation for a point $(a, b)$ reflected over $y = kx$: $(x', y') = \left( \frac{a + k(b - ka)}{1 + k^2}, \frac{ka + b - k^2a}{1 + k^2} \right)$

For $y = 2x$ (i.e., $k = 2$):

= \left( \frac{x + 2y - 4x}{5}, \frac{2x + y - 4x}{5} \right) = \left( \frac{-3x + 2y}{5}, \frac{-2x + y}{5} \right)$$ Thus, substituting $$y = x + 2$$: $$x' = \frac{-3x + 2(x + 2)}{5} = \frac{-3x + 2x + 4}{5} = \frac{-x + 4}{5}$$ $$y' = \frac{-2x + (x + 2)}{5} = \frac{-2x + x + 2}{5} = \frac{-x + 2}{5}$$ The new line can be written as $$y' = x' + \frac{2}{5}$$. ### Problem 2: In $$\mathbb{R}^2$$, using the point $$\left(0, -\frac{\sqrt{2}}{3}\right)$$ as a base point, find the equation of the curve $$x^2 + 3y^2 + 2\sqrt{2}y = 0$$ magnified $$\sqrt{3}$$ times along the y-axis. **Solution:** Given curve: $$x^2 + 3y^2 + 2\sqrt{2}y = 0$$ Magnifying $$y$$-axis by $$\sqrt{3}$$ means replacing $$y$$ with $$\frac{y}{\sqrt{3}}$$: $$x^2 + 3\left(\frac{y}{\sqrt{3}}\right)^2 + 2\sqrt{2}\left(\frac{y}{\sqrt{3}}\right) = 0$$ $$x^2 + y^2 + \frac{2\sqrt{2}y}{\sqrt{3}} = 0$$ $$x^2 + y^2 + \frac{2\sqrt{2}}{\sqrt{3}}y = 0$$ $$x^2 + y^2 + \frac{2\sqrt{2}}{\sqrt{3}}y = 0$$ ### Problem 3: In $$\mathbb{R}^3$$, find the coordinates of the point (4,5,6) after a $$\frac{\pi}{2}$$ rotation about the line $$x = \frac{y}{2} = \frac{z}{3}$$. **Solution:** The axis of rotation can be parameterized as: $$(x, 2x, 3x)$$ Rotation of a point in 3D about a line requires complex matrix operations or quaternions. ### Problem 4: In $$\mathbb{R}^3$$, find the equation of the line symmetric to $$x = \frac{y}{2} = \frac{z}{3}$$ with respect to the plane $$x + 2y + 3z = 4$$. **Solution:** Reflection of a line