Answer :
To solve the problem, let's go through the steps to determine the slope [tex]\( m \)[/tex] and the length of [tex]\(\overline{A'B'}\)[/tex] after the dilation.
### Step 1: Calculate the Slope
First, we need to find the slope [tex]\( m \)[/tex] of the line segment [tex]\(\overline{AB}\)[/tex] with endpoints [tex]\( A(2,2) \)[/tex] and [tex]\( B(3,8) \)[/tex]. The formula for the slope between 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]
Substituting the given points:
[tex]\[ m = \frac{8 - 2}{3 - 2} = \frac{6}{1} = 6 \][/tex]
### Step 2: Calculate the Original Length of [tex]\(\overline{AB}\)[/tex]
Next, we use the distance formula to find the original length of [tex]\(\overline{AB}\)[/tex], given by:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the coordinates of [tex]\( A(2,2) \)[/tex] and [tex]\( B(3,8) \)[/tex]:
[tex]\[ d = \sqrt{(3 - 2)^2 + (8 - 2)^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37} \][/tex]
### Step 3: Calculate the Length of [tex]\(\overline{A'B'}\)[/tex] after Dilation
The line segment [tex]\(\overline{AB}\)[/tex] is dilated by a scale factor of [tex]\( 3.5 \)[/tex]. The new length [tex]\( d' \)[/tex] can be found by multiplying the original length by the scale factor:
[tex]\[ d' = \text{original length} \times \text{scale factor} \][/tex]
[tex]\[ d' = \sqrt{37} \times 3.5 = 3.5 \sqrt{37} \][/tex]
### Step 4: Confirm the Choices
Now, we compare our results with the given answer choices:
- Slope [tex]\( m = 6 \)[/tex]
- Length [tex]\( d' = 3.5 \sqrt{37} \)[/tex]
These match option B:
- [tex]\( m = 6 \)[/tex]
- [tex]\(\overline{A'B'} = 3.5 \sqrt{37} \)[/tex]
Thus, the answer is:
[tex]\[ \boxed{B. \, m=6, \, A^{\prime} B^{\prime}=3.5 \sqrt{37}} \][/tex]
### Step 1: Calculate the Slope
First, we need to find the slope [tex]\( m \)[/tex] of the line segment [tex]\(\overline{AB}\)[/tex] with endpoints [tex]\( A(2,2) \)[/tex] and [tex]\( B(3,8) \)[/tex]. The formula for the slope between 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]
Substituting the given points:
[tex]\[ m = \frac{8 - 2}{3 - 2} = \frac{6}{1} = 6 \][/tex]
### Step 2: Calculate the Original Length of [tex]\(\overline{AB}\)[/tex]
Next, we use the distance formula to find the original length of [tex]\(\overline{AB}\)[/tex], given by:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the coordinates of [tex]\( A(2,2) \)[/tex] and [tex]\( B(3,8) \)[/tex]:
[tex]\[ d = \sqrt{(3 - 2)^2 + (8 - 2)^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37} \][/tex]
### Step 3: Calculate the Length of [tex]\(\overline{A'B'}\)[/tex] after Dilation
The line segment [tex]\(\overline{AB}\)[/tex] is dilated by a scale factor of [tex]\( 3.5 \)[/tex]. The new length [tex]\( d' \)[/tex] can be found by multiplying the original length by the scale factor:
[tex]\[ d' = \text{original length} \times \text{scale factor} \][/tex]
[tex]\[ d' = \sqrt{37} \times 3.5 = 3.5 \sqrt{37} \][/tex]
### Step 4: Confirm the Choices
Now, we compare our results with the given answer choices:
- Slope [tex]\( m = 6 \)[/tex]
- Length [tex]\( d' = 3.5 \sqrt{37} \)[/tex]
These match option B:
- [tex]\( m = 6 \)[/tex]
- [tex]\(\overline{A'B'} = 3.5 \sqrt{37} \)[/tex]
Thus, the answer is:
[tex]\[ \boxed{B. \, m=6, \, A^{\prime} B^{\prime}=3.5 \sqrt{37}} \][/tex]