Given [tex]\triangle ABC[/tex] with [tex]\overline{AB}[/tex] contained in the line [tex]2x + 3y = 5[/tex]. If [tex]\triangle ABC[/tex] is dilated to get [tex]\triangle A'B'C[/tex], which of the following lines could contain [tex]\overline{A'B'}[/tex]?

A. [tex]3x - 2y = 10[/tex]
B. [tex]2x + 3y = 10[/tex]
C. [tex]3x + 2y = 10[/tex]
D. [tex]2x - 3y = 10[/tex]



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]