7) What is the image of the line given by $$ (x, y)=(-1,0)+t h $$ ted, Under the translatron that takes $$ (1,0) $$ to $$ (0,1) $$ followed. the reflection about the line $$ y=2 x $$ ?

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To find the image of the line given by $$ (x, y) = (-1, 0) + t h $$ under the specified transformations, we will follow these steps:

1. **Translate the line**: We will first apply the translation that takes $$ (1, 0) $$ to $$ (0, 1) $$.
2. **Reflect the translated line**: Next, we will reflect the translated line about the line $$ y = 2x $$.

### Step 1: Translation

The translation vector can be determined by finding the difference between the target point and the original point:
$$
\text{Translation vector} = (0, 1) - (1, 0) = (-1, 1)
$$

Now, we apply this translation to the line $$ (x, y) = (-1, 0) + t h $$:
$$
(x', y') = (-1, 0) + (-1, 1) + t h = (-2 + t h_x, 1 + t h_y)
$$
where $$ h = (h_x, h_y) $$.

### Step 2: Reflection about the line $$ y = 2x $$

To reflect a point $$ (x', y') $$ about the line $$ y = 2x $$, we can use the following formula for reflection across a line $$ Ax + By + C = 0 $$:
$$
\text{Reflected point} = \left( \frac{x' (B^2 - A^2) - 2AB y' - 2AC}{A^2 + B^2}, \frac{y' (A^2 - B^2) - 2AB x' - 2BC}{A^2 + B^2} \right)
$$
For the line $$ y = 2x $$, we can rewrite it in the form $$ -2x + y = 0 $$, which gives us $$ A = -2, B = 1, C = 0 $$.

Now, we will apply this reflection formula to the translated point $$ (x', y') $$.

Let's calculate the reflected point using the above formula.

### Calculation

We will perform the calculations for the reflection of the translated line point $$ (x', y') $$ using the reflection formula.

Let’s denote:
- $$ A = -2 $$
- $$ B = 1 $$
- $$ C = 0 $$

Now we will calculate the reflected point.

The reflected coordinates will be:
$$
\text{Reflected } x = \frac{x' (1^2 - (-2)^2) - 2(-2)(1)y' - 2(-2)(0)}{(-2)^2 + 1^2}
$$
$$
\text{Reflected } y = \frac{y' ((-2)^2 - 1^2) - 2(-2)(1)x' - 2(1)(0)}{(-2)^2 + 1^2}
$$

Now, let's compute these values.

### Final Calculation

We will compute the reflection for the general point $$ (x', y') = (-2 + t h_x, 1 + t h_y) $$.

Let's perform the calculations.

1. **Calculate the reflected x-coordinate**:
$$
\text{Reflected } x = \frac{(-2 + t h_x)(1 - 4) + 4(1)(1 + t h_y)}{4 + 1}
$$
$$
= \frac{(-2 + t h_x)(-3) + 4(1 + t h_y)}{5}
$$
$$
= \frac{6 - 3t h_x + 4 + 4t h_y}{5}
$$
$$
= \frac{10 - 3t h_x + 4t h_y}{5}
$$

2. **Calculate the reflected y-coordinate**:
$$
\text{Reflected } y = \frac{(1 + t h_y)(4 - 1) + 4(-2)(-2 + t h_x)}{5}
$$
$$
= \frac{(1 + t h_y)(3) + 16 - 8t h_x}{5}
$$
$$
= \frac{3 + 3t h_y + 16 - 8t h_x}{5}
$$
$$
= \frac{19 + 3t h_y - 8t h_x}{5}
$$

### Conclusion

The image of the line under the specified transformations is given by the parametric equations:
$$
(x'', y'') = \left( \frac{10 - 3t h_x + 4t h_y}{5}, \frac{19 + 3t h_y - 8t h_x}{5} \right)
$$
This represents the transformed line after translation and reflection.
The image of the line after translation and reflection is given by the parametric equations: $$ (x'', y'') = \left( \frac{10 - 3t h_x + 4t h_y}{5}, \frac{19 + 3t h_y - 8t h_x}{5} \right) $$

7) What is the image of the line given by \( (x, y)=(-1,0)+t h \) ted, Under the translatron that takes \( (1,0) \) to \( (0,1) \) followed. the reflection about the line \( y=2 x \) ?

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