line y=3x-1 is transformed by a dilation with a scale factor of 2 and centered at (3,8). The line's image is

Line MN is dilated by a scale factor of 2 centered at the point (0,6). If MN is represented by y=-3x+6, The line's image is

y=-6x+12.

To find the image of the line after a dilation, we need to apply the transformation to the equation of the original line.

For the first question, the line y = 3x - 1 is being dilated with a scale factor of 2 centered at (3,8). A dilation with a scale factor of 2 means that every point on the original line will be multiplied by 2 in both the x and y directions.

To start, we need to shift the center of dilation to the origin (0,0) by subtracting the coordinates of the center from the equation. In this case, we subtract (3,8) from the equation.

So, the new equation becomes:
y = 3(x - 3) - 1 + 8.

Simplify the equation:
y = 3x - 9 - 1 + 8.
y = 3x - 2.

Therefore, the image of the line y = 3x - 1 after a dilation with a scale factor of 2 centered at (3,8) is y = 3x - 2.

For the second question, the line MN with the equation y = -3x + 6 is being dilated with a scale factor of 2 centered at (0,6). Applying the same steps as above, we have:

Shifting the center to the origin (0,0):
y = -3(x - 0) + 6 - 0.

Simplify the equation:
y = -3x + 6.

Therefore, the image of the line y = -3x + 6 after a dilation with a scale factor of 2 centered at (0,6) is y = -3x + 6.

To find the image of the line y=3x-1 when dilated by a scale factor of 2 and centered at (3,8), we can follow these steps:

1. First, we need to find the new coordinates of the center of dilation.
- The original center of dilation is (3,8).
- Since the scale factor is 2, the new center of dilation will be (3,8), as the center remains unchanged.

2. Next, we need to find the image of a single point on the line.
- Choose any point on the original line. Let's pick (0,-1) as an example.
- Apply the dilation to this point using the scale factor of 2 and the new center of dilation:
- (x', y') = (scale factor * (x - center_x) + center_x, scale factor * (y - center_y) + center_y)
- Plugging in the values, we get:
- (x', y') = (2 * (0 - 3) + 3, 2 * (-1 - 8) + 8)
- Simplifying, we get (x', y') = (3, 6)

3. Now, we have the new coordinates of a point on the image line: (3, 6).
- We can use this point and the center of dilation (3, 8) to determine the equation of the image line.
- The equation will have the form: y = mx + b, where m is the slope and b is the y-intercept.
- To find the slope, we can use the formula: m = (y' - center_y) / (x' - center_x)
- Plugging in the values, we get: m = (6 - 8) / (3 - 3) = 0/0 (Undefined)

Since the slope is undefined, we can conclude that the image of the line y=3x-1, under the given transformation, will be a vertical line passing through the point (3, 6).