Polygon [tex]$ABCD$[/tex] is dilated by a scale factor of 2 with the center of dilation at the origin to create polygon [tex]$A'B'C'D'$[/tex]. If the endpoints of [tex]$\overline{AB}$[/tex] are located at [tex]$(0, -7)$[/tex] and [tex]$(8, 8)$[/tex], what is the length of [tex]$\overline{A'B'}$[/tex]?

Use the distance formula to help you decide:
[tex]\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\][/tex]



Answer :

To solve this problem, we need to determine the length of segment [tex]\( \overline{A'B'} \)[/tex] after dilation when the original segment [tex]\( \overline{AB} \)[/tex] is dilated by a scale factor of 2.

First, we start by identifying the coordinates of the endpoints of [tex]\( \overline{AB} \)[/tex]:
- Coordinates of point [tex]\( A \)[/tex]: [tex]\( (0, -7) \)[/tex]
- Coordinates of point [tex]\( B \)[/tex]: [tex]\( (8, 8) \)[/tex]

Next, we need to calculate the length of segment [tex]\( \overline{AB} \)[/tex] using the distance formula:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

Plugging in the coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ d_{AB} = \sqrt{(8 - 0)^2 + (8 - (-7))^2} \][/tex]
[tex]\[ d_{AB} = \sqrt{8^2 + (8 + 7)^2} \][/tex]
[tex]\[ d_{AB} = \sqrt{64 + 15^2} \][/tex]
[tex]\[ d_{AB} = \sqrt{64 + 225} \][/tex]
[tex]\[ d_{AB} = \sqrt{289} \][/tex]
[tex]\[ d_{AB} = 17 \][/tex]

Now, we need to find the length of segment [tex]\( \overline{A'B'} \)[/tex] after the dilation by a scale factor of 2. In a dilation with the center at the origin, each length is scaled by the scale factor. Therefore, the length of [tex]\( \overline{A'B'} \)[/tex] is:
[tex]\[ d_{A'B'} = 2 \times d_{AB} \][/tex]
[tex]\[ d_{A'B'} = 2 \times 17 \][/tex]
[tex]\[ d_{A'B'} = 34 \][/tex]

Therefore, the length of [tex]\( \overline{A'B'} \)[/tex] is [tex]\( 34 \)[/tex].