Answer :
To solve this problem, we need to determine the slope and the length of the segment after dilation.
Let's start by finding the slope (m) of [tex]$\overline{AB}$[/tex].
The slope formula is:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Given 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]\( m = 6 \)[/tex].
Next, we need to determine 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]
Substituting in 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]\[ d = \sqrt{37} \][/tex]
The length of [tex]$\overline{AB}$[/tex] is [tex]\( \sqrt{37} \)[/tex].
Now, [tex]$\overline{A B}$[/tex] is dilated by a scale factor of 3.5 with the origin as the center to get [tex]$\overline{A^{\prime} B^{\prime \prime}}$[/tex]. The length of the dilated segment [tex]$\overline{A^{\prime} B^{\prime \prime}}$[/tex] is scaled by the factor 3.5:
[tex]\[ \text{Length of } \overline{A^{\prime} B^{\prime \prime}} = 3.5 \times \sqrt{37} \][/tex]
Thus, we have:
- Slope [tex]\( m = 6 \)[/tex]
- Length of $\overline{A^{\prime} B^{\prime \prime}} = 3.5 \sqrt{37} \)
So, the correct answer choice is:
D. [tex]\( m=6, A^{\prime} B^{\prime \prime} = 3.5 \sqrt{37} \)[/tex]
Let's start by finding the slope (m) of [tex]$\overline{AB}$[/tex].
The slope formula is:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Given 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]\( m = 6 \)[/tex].
Next, we need to determine 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]
Substituting in 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]\[ d = \sqrt{37} \][/tex]
The length of [tex]$\overline{AB}$[/tex] is [tex]\( \sqrt{37} \)[/tex].
Now, [tex]$\overline{A B}$[/tex] is dilated by a scale factor of 3.5 with the origin as the center to get [tex]$\overline{A^{\prime} B^{\prime \prime}}$[/tex]. The length of the dilated segment [tex]$\overline{A^{\prime} B^{\prime \prime}}$[/tex] is scaled by the factor 3.5:
[tex]\[ \text{Length of } \overline{A^{\prime} B^{\prime \prime}} = 3.5 \times \sqrt{37} \][/tex]
Thus, we have:
- Slope [tex]\( m = 6 \)[/tex]
- Length of $\overline{A^{\prime} B^{\prime \prime}} = 3.5 \sqrt{37} \)
So, the correct answer choice is:
D. [tex]\( m=6, A^{\prime} B^{\prime \prime} = 3.5 \sqrt{37} \)[/tex]
