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 \( y=2 x \) ?

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.