Let's work through this step by step.
### Step 1: Find the Slope of [tex]\(\overline{AB}\)[/tex]
The slope formula is given by:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Plugging in the coordinates [tex]\(A(2, 2)\)[/tex] and [tex]\(B(3, 8)\)[/tex]:
[tex]\[ m = \frac{8 - 2}{3 - 2} = \frac{6}{1} = 6 \][/tex]
So, the slope [tex]\(m\)[/tex] is [tex]\(6.0\)[/tex].
### Step 2: Calculate the Length of [tex]\(\overline{AB}\)[/tex]
Using the distance formula:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Again, using the coordinates [tex]\(A(2, 2)\)[/tex] and [tex]\(B(3, 8)\)[/tex]:
[tex]\[ d = \sqrt{(3 - 2)^2 + (8 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{1^2 + 6^2} \][/tex]
[tex]\[ d = \sqrt{1 + 36} \][/tex]
[tex]\[ d = \sqrt{37} \approx 6.082762530298219 \][/tex]
So, the length of [tex]\(\overline{AB}\)[/tex] is approximately [tex]\(6.082762530298219\)[/tex].
### Step 3: Calculate the Length of [tex]\(\overline{A'B'}\)[/tex] After Dilation
The dilation factor is [tex]\(3.5\)[/tex]. When a line segment is dilated by a scale factor [tex]\(k\)[/tex], the length of the image is [tex]\(k\)[/tex] times the original length.
[tex]\[ \text{Length of } \overline{A'B'} = 3.5 \times \sqrt{37} \][/tex]
Calculating the exact length:
[tex]\[ \text{Length of } \overline{A'B'} = 3.5 \times 6.082762530298219 \approx 21.289668856043768 \][/tex]
So, the length of [tex]\(\overline{A'B'}\)[/tex] after dilation is approximately [tex]\(21.289668856043768\)[/tex].
### Conclusion
From the calculations, we have:
- Slope [tex]\(m = 6.0\)[/tex]
- Length of [tex]\(\overline{A'B'} = 21.289668856043768\)[/tex], which can be represented as [tex]\(3.5 \sqrt{37}\)[/tex] in simplified form.
Hence, the correct answer is:
B. [tex]\(m = 6, A^{\prime} B^{\prime}=3.5 \sqrt{37}\)[/tex]
