To find the slope and length of the dilated line segment [tex]\(\overline{A'B'}\)[/tex] when [tex]\(\overline{A B}\)[/tex] is dilated by a scale factor of 3.5 with the origin as the center of dilation, we can follow these detailed steps.
### Step 1: Find the Slope [tex]\((m)\)[/tex] of [tex]\(\overline{A B}\)[/tex]
The slope [tex]\(m\)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is calculated using the formula:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Given the endpoints [tex]\(A(2, 2)\)[/tex] and [tex]\(B(3, 8)\)[/tex], we can find the slope:
[tex]\[ m = \frac{8 - 2}{3 - 2} = \frac{6}{1} = 6 \][/tex]
So, the slope [tex]\(m\)[/tex] is [tex]\(6\)[/tex].
### Step 2: Find the Length of [tex]\(\overline{A B}\)[/tex]
The length [tex]\(d\)[/tex] of a line segment between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by the distance formula:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Using the given points [tex]\(A(2, 2)\)[/tex] and [tex]\(B(3, 8)\)[/tex], we can find the length:
[tex]\[ d = \sqrt{(3 - 2)^2 + (8 - 2)^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37} \][/tex]
So, the length of [tex]\(\overline{A B}\)[/tex] is [tex]\(\sqrt{37}\)[/tex].
### Step 3: Calculate the Length of [tex]\(\overline{A^{\prime} B^{\prime}}\)[/tex]
Given that [tex]\(\overline{A B}\)[/tex] is dilated by a scale factor of 3.5, the length of [tex]\(\overline{A^{\prime} B^{\prime}}\)[/tex] will be:
[tex]\[ \text{length of } \overline{A^{\prime} B^{\prime}} = 3.5 \times \sqrt{37} \][/tex]
### Summarize Results
- The slope [tex]\(m\)[/tex] remains the same after dilation, so [tex]\(m = 6\)[/tex].
- The length of [tex]\(\overline{A'B'}\)[/tex] is [tex]\(3.5 \sqrt{37}\)[/tex].
Thus, the correct choice is:
[tex]\[ \boxed{D. \, m=6, \, A^{\prime} B^{\prime}=3.5 \sqrt{37}} \][/tex]
