https://file.aiquickdraw.com/mathbot/file/1727306510970z7pfuavk.jpg

Let's analyze the image and questions.

In the graph, triangle \(ABC\) has the following coordinates:

- Point \(A(2, 0)\),
- Point \(B(4, 1)\),
- Point \(C(6, 3)\).

The dilation is centered at \( (2, 0) \) and we are asked to scale the triangle by different factors. Here's how the dilation works:

### a. Dilation of triangle ABC with center \((2, 0)\) and scale factor 2:
To perform dilation, we apply the formula:
\[
\text{New coordinates} = (\text{Center coordinates}) + \text{scale factor} \times (\text{Original coordinates} - \text{Center coordinates})
\]
For each point:

- Point \(A(2, 0)\) stays at \(A'(2, 0)\) (since it’s the center of dilation).
- Point \(B(4, 1)\):
  \[
  B' = (2, 0) + 2 \times (4 - 2, 1 - 0) = (2, 0) + 2 \times (2, 1) = (2 + 4, 0 + 2) = (6, 2)
  \]
- Point \(C(6, 3)\):
  \[
  C' = (2, 0) + 2 \times (6 - 2, 3 - 0) = (2, 0) + 2 \times (4, 3) = (2 + 8, 0 + 6) = (10, 6)
  \]

### b. Dilation of triangle ABC with center \((2, 0)\) and scale factor 3:
- Point \(A(2, 0)\) stays at \(A'(2, 0)\).
- Point \(B(4, 1)\):
  \[
  B' = (2, 0) + 3 \times (4 - 2, 1 - 0) = (2, 0) + 3 \times (2, 1) = (8, 3)
  \]
- Point \(C(6, 3)\):
  \[
  C' = (2, 0) + 3 \times (6 - 2, 3 - 0) = (2, 0) + 3 \times (4, 3) = (14, 9)
  \]

### c. Dilation of triangle ABC with center \((2, 0)\) and scale factor \(\frac{1}{2}\):
- Point \(A(2, 0)\) stays at \(A'(2, 0)\).
- Point \(B(4, 1)\):
  \[
  B' = (2, 0) + \frac{1}{2} \times (4 - 2, 1 - 0) = (2, 0) + \frac{1}{2} \times (2, 1) = (3, 0.5)
  \]
- Point \(C(6, 3)\):
  \[
  C' = (2, 0) + \frac{1}{2} \times (6 - 2, 3 - 0) = (2, 0) + \frac{1}{2} \times (4, 3) = (4, 1.5)
  \]

### d. Coordinates of point \(C\) for a dilation with scale factor \(s\):
The formula is the same as above. For point \(C(6, 3)\), after dilation:
\[
C' = (2, 0) + s \times (6 - 2, 3 - 0) = (2, 0) + s \times (4, 3) = (2 + 4s, 3s)
\]
Thus, the coordinates of the image of \(C\) are \((2 + 4s, 3s)\).

### e. Equation of the line containing all possible images of point \(C\):
From part d, we know that the coordinates of the dilated point \(C\) are \((2 + 4s, 3s)\). To find the equation of the line, we eliminate the parameter \(s\).

Since \(x = 2 + 4s\) and \(y = 3s\), we solve for \(s\) in terms of \(x\):
\[
s = \frac{x - 2}{4}
\]
Substitute this into the equation for \(y\):
\[
y = 3 \times \frac{x - 2}{4} = \frac{3(x - 2)}{4}
\]
Thus, the equation of the line is:
\[
y = \frac{3}{4}(x - 2)
\]

Do you have any questions or want more details?

Here are some related questions:
1. What happens to the line if the center of dilation changes?
2. How does the scale factor affect the slope of the line?
3. How would the dilation work if the center of dilation were not on the x-axis?
4. What is the equation of the line containing all possible images of point \(B\)?
5. Can you dilate a shape by a negative scale factor? What would that look like?

**Tip:** When working with dilations, always pay attention to the center of dilation, as it determines how far each point moves.

Draw the dilation of triangle ABC with center (2, 0) and various scale factors, and write an equation for the line containing all possible images of point C.

Learn how to perform a dilation of triangle ABC with center (2, 0) and various scale factors. Find new coordinates and write the equation for the line containing all possible images of point C. This problem covers key concepts in coordinate geometry and transformations, suitable for students in Grades 8-10.

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