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The endpoints of [tex]\overline{AB}[/tex] are [tex]\(A(2,2)\)[/tex] and [tex]\(B(3,8)\)[/tex]. [tex]\overline{AB}[/tex] is dilated by a scale factor of 3.5 with the origin as the center of dilation to give the image [tex]\overline{A'B'}[/tex].

What are the slope [tex]\((m)\)[/tex] and 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]

A. [tex]\(m=21, A'B' = \sqrt{37}\)[/tex]

B. [tex]\(m=6, A'B' = 3.5\sqrt{37}\)[/tex]

C. [tex]\(m=6, A'B' = \sqrt{37}\)[/tex]

D. [tex]\(m=21, A'B' = 3.5\sqrt{37}\)[/tex]



Answer :

GinnyAnswer
Let's analyze the problem step-by-step:

1. Identify the coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
- [tex]\( A = (2, 2) \)[/tex]
- [tex]\( B = (3, 8) \)[/tex]

2. Calculate the slope [tex]\( m \)[/tex] of [tex]\(\overline{AB}\)[/tex]:
- The slope formula is [tex]\( m = \frac{y_2 - y_1}{x_2 - x_1} \)[/tex]
- Substituting the coordinates of the points:
[tex]\[ m = \frac{8 - 2}{3 - 2} = \frac{6}{1} = 6 \][/tex]

3. Calculate the original 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 the coordinates of the points:
[tex]\[ d = \sqrt{(3 - 2)^2 + (8 - 2)^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37} \][/tex]

4. Determine the new length of [tex]\(\overline{A'B'}\)[/tex] after dilation:
- The dilation scale factor is 3.5.
- The length of [tex]\(\overline{A'B'}\)[/tex] is the original length multiplied by the scale factor:
[tex]\[ L_{A'B'} = 3.5 \times \sqrt{37} \][/tex]

5. Combine the results:
- The slope [tex]\( m \)[/tex] remains unchanged after dilation.
- The new length of [tex]\(\overline{A'B'}\)[/tex] is [tex]\( 3.5 \sqrt{37} \)[/tex].

Putting it all together, the final values are:
- Slope [tex]\( m = 6 \)[/tex]
- Length of [tex]\(\overline{A'B'} = 3.5 \sqrt{37} \)[/tex]

Therefore, the correct answer is:
B. [tex]\( m = 6 \)[/tex], [tex]\( A'B' = 3.5 \sqrt{37} \)[/tex].

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