Let's solve this problem step by step:
1. Finding the Slope of Line Segment [tex]\(\overline{AB}\)[/tex]:
- The coordinates of point [tex]\(A\)[/tex] are [tex]\((2, 2)\)[/tex] and the coordinates of point [tex]\(B\)[/tex] are [tex]\((3, 8)\)[/tex].
- To find the slope [tex]\(m\)[/tex] of line segment [tex]\(\overline{AB}\)[/tex], we use the slope formula:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
Substituting the given coordinates:
[tex]\[
m = \frac{8 - 2}{3 - 2} = \frac{6}{1} = 6
\][/tex]
2. Finding the Length of [tex]\(\overline{AB}\)[/tex] Using the Distance Formula:
- The distance formula is given by:
[tex]\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\][/tex]
Substituting the coordinates of points [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[
d = \sqrt{(3 - 2)^2 + (8 - 2)^2} = \sqrt{1 + 36} = \sqrt{37}
\][/tex]
The length of [tex]\(\overline{AB}\)[/tex] is [tex]\(\sqrt{37}\)[/tex].
3. Finding the Length of the Dilated Line Segment [tex]\(\overline{A'B'}\)[/tex]:
- The dilation scale factor is given as [tex]\(3.5\)[/tex].
- To find the length of [tex]\(\overline{A'B'}\)[/tex], we multiply the original length of [tex]\(\overline{AB}\)[/tex] by the scale factor:
[tex]\[
\text{Length of } \overline{A'B'} = 3.5 \times \sqrt{37}
\][/tex]
Now that we have all the required components:
- The slope [tex]\(m\)[/tex] of [tex]\(\overline{AB}\)[/tex] is [tex]\(6\)[/tex].
- The length of the dilated line segment [tex]\(\overline{A'B'}\)[/tex] is [tex]\(3.5 \sqrt{37}\)[/tex].
Thus, the correct answer is:
C. [tex]\(m = 6, A^{\prime}B^{\prime} = 3.5 \sqrt{37}\)[/tex]
