Answer :
To solve this problem, we need to identify which one of the given lines could contain the segment [tex]\(\overline{A'B'}\)[/tex], after the dilation of [tex]\(\triangle ABC\)[/tex].
First, let's analyze the original line given for [tex]\(\overline{AB}\)[/tex]:
[tex]\[ 2x + 3y = 5 \][/tex]
When a geometric shape undergoes dilation, the parallelism of lines remains preserved. This means that the line containing [tex]\(\overline{A'B'}\)[/tex] must be parallel to the original line containing [tex]\(\overline{AB}\)[/tex].
For a line to be parallel to another, their equations must have proportional coefficients for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
Given the following options, we need to identify which one has the same slope as the line [tex]\(2x + 3y = 5\)[/tex]:
1. [tex]\(3x - 2y = 10\)[/tex]
2. [tex]\(2x + 3y = 10\)[/tex]
3. [tex]\(3x + 2y = 10\)[/tex]
4. [tex]\(2x - 3y = 10\)[/tex]
The slope of a line in the form [tex]\(Ax + By = C\)[/tex] is given by [tex]\(-\frac{A}{B}\)[/tex]. For the given line [tex]\(2x + 3y = 5\)[/tex], the slope is [tex]\(-\frac{2}{3}\)[/tex].
Let's check the slope of each option:
1. For [tex]\(3x - 2y = 10\)[/tex]:
- The slope is [tex]\(-\frac{3}{-2} = \frac{3}{2}\)[/tex], which is not equal to [tex]\(-\frac{2}{3}\)[/tex].
2. For [tex]\(2x + 3y = 10\)[/tex]:
- The slope is [tex]\(-\frac{2}{3}\)[/tex], which matches the slope of the given line [tex]\(2x + 3y = 5\)[/tex].
3. For [tex]\(3x + 2y = 10\)[/tex]:
- The slope is [tex]\(-\frac{3}{2}\)[/tex], not equal to [tex]\(-\frac{2}{3}\)[/tex].
4. For [tex]\(2x - 3y = 10\)[/tex]:
- The slope is [tex]\(-\frac{2}{-3} = \frac{2}{3}\)[/tex], which also is not equal to [tex]\(-\frac{2}{3}\)[/tex].
From the analysis, option 2 [tex]\(2x + 3y = 10\)[/tex] has the coefficients proportional to the original line [tex]\(2x + 3y = 5\)[/tex]. Therefore, it is parallel and hence could contain the segment [tex]\(\overline{A'B'}\)[/tex] after the dilation.
So, the correct line that could contain [tex]\(\overline{A'B'}\)[/tex] is:
[tex]\[ 2x + 3y = 10 \][/tex]
Thus, the answer is:
[tex]\[ \boxed{2} \][/tex]
First, let's analyze the original line given for [tex]\(\overline{AB}\)[/tex]:
[tex]\[ 2x + 3y = 5 \][/tex]
When a geometric shape undergoes dilation, the parallelism of lines remains preserved. This means that the line containing [tex]\(\overline{A'B'}\)[/tex] must be parallel to the original line containing [tex]\(\overline{AB}\)[/tex].
For a line to be parallel to another, their equations must have proportional coefficients for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
Given the following options, we need to identify which one has the same slope as the line [tex]\(2x + 3y = 5\)[/tex]:
1. [tex]\(3x - 2y = 10\)[/tex]
2. [tex]\(2x + 3y = 10\)[/tex]
3. [tex]\(3x + 2y = 10\)[/tex]
4. [tex]\(2x - 3y = 10\)[/tex]
The slope of a line in the form [tex]\(Ax + By = C\)[/tex] is given by [tex]\(-\frac{A}{B}\)[/tex]. For the given line [tex]\(2x + 3y = 5\)[/tex], the slope is [tex]\(-\frac{2}{3}\)[/tex].
Let's check the slope of each option:
1. For [tex]\(3x - 2y = 10\)[/tex]:
- The slope is [tex]\(-\frac{3}{-2} = \frac{3}{2}\)[/tex], which is not equal to [tex]\(-\frac{2}{3}\)[/tex].
2. For [tex]\(2x + 3y = 10\)[/tex]:
- The slope is [tex]\(-\frac{2}{3}\)[/tex], which matches the slope of the given line [tex]\(2x + 3y = 5\)[/tex].
3. For [tex]\(3x + 2y = 10\)[/tex]:
- The slope is [tex]\(-\frac{3}{2}\)[/tex], not equal to [tex]\(-\frac{2}{3}\)[/tex].
4. For [tex]\(2x - 3y = 10\)[/tex]:
- The slope is [tex]\(-\frac{2}{-3} = \frac{2}{3}\)[/tex], which also is not equal to [tex]\(-\frac{2}{3}\)[/tex].
From the analysis, option 2 [tex]\(2x + 3y = 10\)[/tex] has the coefficients proportional to the original line [tex]\(2x + 3y = 5\)[/tex]. Therefore, it is parallel and hence could contain the segment [tex]\(\overline{A'B'}\)[/tex] after the dilation.
So, the correct line that could contain [tex]\(\overline{A'B'}\)[/tex] is:
[tex]\[ 2x + 3y = 10 \][/tex]
Thus, the answer is:
[tex]\[ \boxed{2} \][/tex]