Answer :
Certainly! Let's determine which pairs of points form lines that are perpendicular to a line with a slope of [tex]\(-\frac{4}{5}\)[/tex]. To do this, we first need to know how perpendicular slopes relate to each other. If two lines are perpendicular, the product of their slopes is [tex]\(-1\)[/tex].
Given the slope of the original line:
[tex]\[ m = -\frac{4}{5} \][/tex]
The slope of the line perpendicular to this slope, [tex]\( m_\perp \)[/tex], can be found using the relationship:
[tex]\[ m \cdot m_\perp = -1 \][/tex]
Substituting [tex]\( m = -\frac{4}{5} \)[/tex] into the equation:
[tex]\[ -\frac{4}{5} \cdot m_\perp = -1 \][/tex]
[tex]\[ m_\perp = \frac{5}{4} \][/tex]
Now, we will calculate the slopes of the lines passing through each pair of points and check if any of them has a slope of [tex]\(\frac{5}{4}\)[/tex].
### Pair 1: [tex]\((-2,0)\)[/tex] and [tex]\((2,5)\)[/tex]
The slope between these two points, [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex], is:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 0}{2 - (-2)} = \frac{5}{4} \][/tex]
### Pair 2: [tex]\((1,5)\)[/tex] and [tex]\((4,-5)\)[/tex]
[tex]\[ \text{slope} = \frac{-5 - 5}{4 - 1} = \frac{-10}{3} = -\frac{10}{3} \][/tex]
### Pair 3: [tex]\((-3,4)\)[/tex] and [tex]\((2,0)\)[/tex]
[tex]\[ \text{slope} = \frac{0 - 4}{2 - (-3)} = \frac{-4}{5} = -\frac{4}{5} \][/tex]
### Pair 4: [tex]\((1,-1)\)[/tex] and [tex]\((6,-5)\)[/tex]
[tex]\[ \text{slope} = \frac{-5 - (-1)}{6 - 1} = \frac{-5 + 1}{6 - 1} = \frac{-4}{5} = -\frac{4}{5} \][/tex]
### Pair 5: [tex]\((2,-1)\)[/tex] and [tex]\((10,9)\)[/tex]
[tex]\[ \text{slope} = \frac{9 - (-1)}{10 - 2} = \frac{9 + 1}{10 - 2} = \frac{10}{8} = \frac{5}{4} \][/tex]
Comparing these slopes with [tex]\(\frac{5}{4}\)[/tex], we find:
- Pair 1: [tex]\((-2,0)\)[/tex] and [tex]\((2,5)\)[/tex] has a slope of [tex]\(\frac{5}{4}\)[/tex]
- Pair 2: [tex]\((1,5)\)[/tex] and [tex]\((4,-5)\)[/tex] does not have a slope of [tex]\(\frac{5}{4}\)[/tex]
- Pair 3: [tex]\((-3,4)\)[/tex] and [tex]\((2,0)\)[/tex] does not have a slope of [tex]\(\frac{5}{4}\)[/tex]
- Pair 4: [tex]\((1,-1)\)[/tex] and [tex]\((6,-5)\)[/tex] does not have a slope of [tex]\(\frac{5}{4}\)[/tex]
- Pair 5: [tex]\((2,-1)\)[/tex] and [tex]\((10,9)\)[/tex] has a slope of [tex]\(\frac{5}{4}\)[/tex]
Hence, the ordered pairs that could be points on a line perpendicular to the given line are:
[tex]\[ (-2, 0) \text{ and } (2, 5) \][/tex]
[tex]\[ (2, -1) \text{ and } (10, 9) \][/tex]
Given the slope of the original line:
[tex]\[ m = -\frac{4}{5} \][/tex]
The slope of the line perpendicular to this slope, [tex]\( m_\perp \)[/tex], can be found using the relationship:
[tex]\[ m \cdot m_\perp = -1 \][/tex]
Substituting [tex]\( m = -\frac{4}{5} \)[/tex] into the equation:
[tex]\[ -\frac{4}{5} \cdot m_\perp = -1 \][/tex]
[tex]\[ m_\perp = \frac{5}{4} \][/tex]
Now, we will calculate the slopes of the lines passing through each pair of points and check if any of them has a slope of [tex]\(\frac{5}{4}\)[/tex].
### Pair 1: [tex]\((-2,0)\)[/tex] and [tex]\((2,5)\)[/tex]
The slope between these two points, [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex], is:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 0}{2 - (-2)} = \frac{5}{4} \][/tex]
### Pair 2: [tex]\((1,5)\)[/tex] and [tex]\((4,-5)\)[/tex]
[tex]\[ \text{slope} = \frac{-5 - 5}{4 - 1} = \frac{-10}{3} = -\frac{10}{3} \][/tex]
### Pair 3: [tex]\((-3,4)\)[/tex] and [tex]\((2,0)\)[/tex]
[tex]\[ \text{slope} = \frac{0 - 4}{2 - (-3)} = \frac{-4}{5} = -\frac{4}{5} \][/tex]
### Pair 4: [tex]\((1,-1)\)[/tex] and [tex]\((6,-5)\)[/tex]
[tex]\[ \text{slope} = \frac{-5 - (-1)}{6 - 1} = \frac{-5 + 1}{6 - 1} = \frac{-4}{5} = -\frac{4}{5} \][/tex]
### Pair 5: [tex]\((2,-1)\)[/tex] and [tex]\((10,9)\)[/tex]
[tex]\[ \text{slope} = \frac{9 - (-1)}{10 - 2} = \frac{9 + 1}{10 - 2} = \frac{10}{8} = \frac{5}{4} \][/tex]
Comparing these slopes with [tex]\(\frac{5}{4}\)[/tex], we find:
- Pair 1: [tex]\((-2,0)\)[/tex] and [tex]\((2,5)\)[/tex] has a slope of [tex]\(\frac{5}{4}\)[/tex]
- Pair 2: [tex]\((1,5)\)[/tex] and [tex]\((4,-5)\)[/tex] does not have a slope of [tex]\(\frac{5}{4}\)[/tex]
- Pair 3: [tex]\((-3,4)\)[/tex] and [tex]\((2,0)\)[/tex] does not have a slope of [tex]\(\frac{5}{4}\)[/tex]
- Pair 4: [tex]\((1,-1)\)[/tex] and [tex]\((6,-5)\)[/tex] does not have a slope of [tex]\(\frac{5}{4}\)[/tex]
- Pair 5: [tex]\((2,-1)\)[/tex] and [tex]\((10,9)\)[/tex] has a slope of [tex]\(\frac{5}{4}\)[/tex]
Hence, the ordered pairs that could be points on a line perpendicular to the given line are:
[tex]\[ (-2, 0) \text{ and } (2, 5) \][/tex]
[tex]\[ (2, -1) \text{ and } (10, 9) \][/tex]
