To solve this problem, we need to find the slope [tex]\( m \)[/tex] and the length of the line segment [tex]\( \overline{A'B'} \)[/tex], which is the dilated image of [tex]\( \overline{AB} \)[/tex].
### Step 1: Calculate the Slope [tex]\( m \)[/tex] of Line Segment [tex]\( \overline{AB} \)[/tex]
Given points:
- [tex]\( A(2, 2) \)[/tex]
- [tex]\( B(3, 8) \)[/tex]
The formula for the slope [tex]\( m \)[/tex] of the line passing through two points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substitute the given coordinates:
[tex]\[ m = \frac{8 - 2}{3 - 2} = \frac{6}{1} = 6 \][/tex]
So, the slope [tex]\( m \)[/tex] is [tex]\( 6 \)[/tex].
### Step 2: Calculate the Length of [tex]\( \overline{AB} \)[/tex] Using the Distance Formula
The distance formula between two points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substitute the given coordinates:
[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]\( \overline{AB} \)[/tex] is [tex]\( \sqrt{37} \)[/tex].
### Step 3: Determine the Length of [tex]\( \overline{A'B'} \)[/tex] After Dilation
The problem specifies a dilation with a scale factor of 3.5. To find the new length of [tex]\( \overline{A'B'} \)[/tex], we multiply the original length [tex]\( \overline{AB} \)[/tex] by the scale factor:
[tex]\[ \text{Length of } \overline{A'B'} = \text{Length of } \overline{AB} \times \text{Scale Factor} \][/tex]
Substitute the known values:
[tex]\[ \text{Length of } \overline{A'B'} = \sqrt{37} \times 3.5 = 3.5\sqrt{37} \][/tex]
Thus, the length of [tex]\( \overline{A'B'} \)[/tex] is [tex]\( 3.5\sqrt{37} \)[/tex].
### Conclusion
From the above steps, we have determined that the slope [tex]\( m \)[/tex] of [tex]\( \overline{A'B'} \)[/tex] is [tex]\( 6 \)[/tex] and the length of [tex]\( \overline{A'B'} \)[/tex] is [tex]\( 3.5\sqrt{37} \)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{B. \; m=6, \; A'B' = 3.5\sqrt{37}} \][/tex]
