Question

Use the image to answer the question.

A coordinate plane with four quadrants shows x-axis ranging from negative 5 to 12 in increments of 1 and the y-axis from 17 to negative 4 in increments of 1. Points upper A, upper B, upper A prime, and upper B prime are plotted on quadrant 1. The plotted points are as follows: Upper A is left parenthesis 2 comma 3 right parenthesis and upper B is left parenthesis 4 comma 1 right parenthesis. The points are joined by a line. Upper A prime is plotted as left parenthesis 6 comma 9 right parenthesis and upper B prime as left parenthesis 12 comma 3 right parenthesis. The points are joined by a line.

Determine the scale factor used in the dilation of line segment AB
, centered at the origin.

(1 point)
The scale factor is
.
Remaining Attempts : 3

Answer 1

To determine the scale factor used in the dilation of line segment \( AB \) centered at the origin, we can follow these steps:

1. **Identify the coordinates of points A and B:**
- \( A(2, 3) \)
- \( B(4, 1) \)

2. **Find the distance between points A and B:**
- The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
- Plugging in the coordinates:
\[
d_{AB} = \sqrt{(4 - 2)^2 + (1 - 3)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]

3. **Identify the coordinates of points \( A' \) and \( B' \):**
- \( A'(6, 9) \)
- \( B'(12, 3) \)

4. **Find the distance between points \( A' \) and \( B' \):**
- Using the same distance formula:
\[
d_{A'B'} = \sqrt{(12 - 6)^2 + (3 - 9)^2} = \sqrt{(6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}
\]

5. **Calculate the scale factor:**
- The scale factor \( k \) can be found by dividing the distance \( A'B' \) by the distance \( AB \):
\[
k = \frac{d_{A'B'}}{d_{AB}} = \frac{6\sqrt{2}}{2\sqrt{2}} = 3
\]

Thus, the scale factor used in the dilation of line segment \( AB \), centered at the origin, is **3**.