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To determine which pairs of points could lie on a line parallel to the line that contains the points [tex]\((3,4)\)[/tex] and [tex]\((-2,2)\)[/tex], we need to compare the slopes of each line. Two lines are parallel if and only if their slopes are equal.
Given points [tex]\((3,4)\)[/tex] and [tex]\((-2,2)\)[/tex], we can calculate the slope using the slope formula:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substitute the values:
[tex]\[ \text{slope} = \frac{2 - 4}{-2 - 3} = \frac{-2}{-5} = \frac{2}{5} \][/tex]
The slope of the line passing through [tex]\((3,4)\)[/tex] and [tex]\((-2,2)\)[/tex] is [tex]\(\frac{2}{5}\)[/tex].
Now, we’ll calculate the slopes of lines corresponding to each of the given pairs and compare them to [tex]\(\frac{2}{5}\)[/tex]:
1. For the points [tex]\( (-2, -5) \)[/tex] and [tex]\( (-7, -3) \)[/tex]:
[tex]\[ \text{slope} = \frac{-3 - (-5)}{-7 - (-2)} = \frac{-3 + 5}{-7 + 2} = \frac{2}{-5} = -\frac{2}{5} \][/tex]
The slope is [tex]\(-\frac{2}{5}\)[/tex], which is not equal to [tex]\(\frac{2}{5}\)[/tex]. Thus, this pair does not lie on a parallel line.
2. For the points [tex]\( (-1, 1) \)[/tex] and [tex]\( (-6, -1) \)[/tex]:
[tex]\[ \text{slope} = \frac{-1 - 1}{-6 - (-1)} = \frac{-1 - 1}{-6 + 1} = \frac{-2}{-5} = \frac{2}{5} \][/tex]
The slope is [tex]\(\frac{2}{5}\)[/tex], which is equal to [tex]\(\frac{2}{5}\)[/tex]. Thus, this pair lies on a parallel line.
3. For the points [tex]\( (0, 0) \)[/tex] and [tex]\( (2, 5) \)[/tex]:
[tex]\[ \text{slope} = \frac{5 - 0}{2 - 0} = \frac{5}{2} \][/tex]
The slope is [tex]\(\frac{5}{2}\)[/tex], which is not equal to [tex]\(\frac{2}{5}\)[/tex]. Thus, this pair does not lie on a parallel line.
4. For the points [tex]\( (1, 0) \)[/tex] and [tex]\( (6, 2) \)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 0}{6 - 1} = \frac{2}{5} \][/tex]
The slope is [tex]\(\frac{2}{5}\)[/tex], which is equal to [tex]\(\frac{2}{5}\)[/tex]. Thus, this pair lies on a parallel line.
5. For the points [tex]\( (3, 0) \)[/tex] and [tex]\( (8, 2) \)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 0}{8 - 3} = \frac{2}{5} \][/tex]
The slope is [tex]\(\frac{2}{5}\)[/tex], which is equal to [tex]\(\frac{2}{5}\)[/tex]. Thus, this pair lies on a parallel line.
Therefore, the ordered pairs that could be points on a line parallel to the line containing [tex]\((3,4)\)[/tex] and [tex]\((-2,2)\)[/tex] are:
[tex]\[(-1,1) \text{ and } (-6,-1)\][/tex]
[tex]\[(1,0) \text{ and } (6,2)\][/tex]
[tex]\[(3, 0) \text{ and } (8, 2)\][/tex]
Given points [tex]\((3,4)\)[/tex] and [tex]\((-2,2)\)[/tex], we can calculate the slope using the slope formula:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substitute the values:
[tex]\[ \text{slope} = \frac{2 - 4}{-2 - 3} = \frac{-2}{-5} = \frac{2}{5} \][/tex]
The slope of the line passing through [tex]\((3,4)\)[/tex] and [tex]\((-2,2)\)[/tex] is [tex]\(\frac{2}{5}\)[/tex].
Now, we’ll calculate the slopes of lines corresponding to each of the given pairs and compare them to [tex]\(\frac{2}{5}\)[/tex]:
1. For the points [tex]\( (-2, -5) \)[/tex] and [tex]\( (-7, -3) \)[/tex]:
[tex]\[ \text{slope} = \frac{-3 - (-5)}{-7 - (-2)} = \frac{-3 + 5}{-7 + 2} = \frac{2}{-5} = -\frac{2}{5} \][/tex]
The slope is [tex]\(-\frac{2}{5}\)[/tex], which is not equal to [tex]\(\frac{2}{5}\)[/tex]. Thus, this pair does not lie on a parallel line.
2. For the points [tex]\( (-1, 1) \)[/tex] and [tex]\( (-6, -1) \)[/tex]:
[tex]\[ \text{slope} = \frac{-1 - 1}{-6 - (-1)} = \frac{-1 - 1}{-6 + 1} = \frac{-2}{-5} = \frac{2}{5} \][/tex]
The slope is [tex]\(\frac{2}{5}\)[/tex], which is equal to [tex]\(\frac{2}{5}\)[/tex]. Thus, this pair lies on a parallel line.
3. For the points [tex]\( (0, 0) \)[/tex] and [tex]\( (2, 5) \)[/tex]:
[tex]\[ \text{slope} = \frac{5 - 0}{2 - 0} = \frac{5}{2} \][/tex]
The slope is [tex]\(\frac{5}{2}\)[/tex], which is not equal to [tex]\(\frac{2}{5}\)[/tex]. Thus, this pair does not lie on a parallel line.
4. For the points [tex]\( (1, 0) \)[/tex] and [tex]\( (6, 2) \)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 0}{6 - 1} = \frac{2}{5} \][/tex]
The slope is [tex]\(\frac{2}{5}\)[/tex], which is equal to [tex]\(\frac{2}{5}\)[/tex]. Thus, this pair lies on a parallel line.
5. For the points [tex]\( (3, 0) \)[/tex] and [tex]\( (8, 2) \)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 0}{8 - 3} = \frac{2}{5} \][/tex]
The slope is [tex]\(\frac{2}{5}\)[/tex], which is equal to [tex]\(\frac{2}{5}\)[/tex]. Thus, this pair lies on a parallel line.
Therefore, the ordered pairs that could be points on a line parallel to the line containing [tex]\((3,4)\)[/tex] and [tex]\((-2,2)\)[/tex] are:
[tex]\[(-1,1) \text{ and } (-6,-1)\][/tex]
[tex]\[(1,0) \text{ and } (6,2)\][/tex]
[tex]\[(3, 0) \text{ and } (8, 2)\][/tex]
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