Answer :
To determine which line is perpendicular to a given line and has the same [tex]\(y\)[/tex]-intercept, we need to understand two key concepts:
1. Perpendicular Slopes: Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex]. If one line has a slope [tex]\(m\)[/tex], the perpendicular line must have a slope [tex]\(m_{\perp}\)[/tex] such that [tex]\(m \cdot m_{\perp} = -1\)[/tex].
2. Same [tex]\(y\)[/tex]-Intercept: The [tex]\(y\)[/tex]-intercept of a line is the point where it crosses the [tex]\(y\)[/tex]-axis. For a line to have the same [tex]\(y\)[/tex]-intercept, it must cross the [tex]\(y\)[/tex]-axis at the same point as the given line.
Given:
- The slope of the original line is [tex]\(-\frac{3}{2}\)[/tex] and it contains the point [tex]\((0, 2)\)[/tex], which indicates the [tex]\(y\)[/tex]-intercept is [tex]\(2\)[/tex].
Let's consider each given option to find out which one meets both conditions.
### Option 1: Slope [tex]$-\frac{3}{2}$[/tex], [tex]\(y\)[/tex]-intercept at [tex]$(0, 2)$[/tex].
- Slope = [tex]$-\frac{3}{2}$[/tex].
- The line passes through the point [tex]$(0, 2)$[/tex] so the [tex]\(y\)[/tex]-intercept is [tex]$2$[/tex].
This line has the correct [tex]\(y\)[/tex]-intercept, but the slope is not the negative reciprocal of the given slope [tex]\( -\frac{3}{2} \)[/tex]. Therefore, it is not perpendicular.
### Option 2: Slope [tex]$-\frac{2}{3}$[/tex], [tex]\(y\)[/tex]-intercept at [tex]$(0, -2)$[/tex].
- Slope = [tex]$-\frac{2}{3}$[/tex].
- The line passes through the point [tex]$(0, -2)$[/tex] so the [tex]\(y\)[/tex]-intercept is [tex]$-2$[/tex].
This [tex]\(y\)[/tex]-intercept is different from [tex]$2$[/tex], and the slope also does not satisfy the perpendicular condition. Thus, this option does not meet our criteria.
### Option 3: Slope [tex]$\frac{3}{2}$[/tex], [tex]\(y\)[/tex]-intercept at [tex]$(0, 2)$[/tex].
- Slope = [tex]$\frac{3}{2}$[/tex].
- The line passes through the point [tex]$(0, 2)$[/tex] so the [tex]\(y\)[/tex]-intercept is [tex]$2$[/tex].
This line has the correct [tex]\(y\)[/tex]-intercept, but for perpendicular slopes, the product with the given slope [tex]$\left(-\frac{3}{2}\right) \cdot \frac{3}{2} = -\frac{9}{4}$[/tex], which is not [tex]\(-1\)[/tex]. Hence, it is not perpendicular.
### Option 4: Slope [tex]$-\frac{3}{2}$[/tex], [tex]\(y\)[/tex]-intercept at [tex]$(0, -2)$[/tex].
- Slope = [tex]$-\frac{3}{2}$[/tex].
- The line passes through the point [tex]$(0, -2)$[/tex] so the [tex]\(y\)[/tex]-intercept is [tex]$-2$[/tex].
This [tex]\(y\)[/tex]-intercept is different from [tex]$2$[/tex], and the slope also does not satisfy the perpendicular condition. Thus, this option does not meet our criteria.
### Correct Criteria
According to our criteria:
- The slope of the perpendicular line must be [tex]$0.6666666666666666$[/tex].
- The [tex]\(y\)[/tex]-intercept must be [tex]$2$[/tex].
The slope that satisfies the perpendicular condition to [tex]\(-\frac{3}{2}\)[/tex] is indeed [tex]$\frac{2}{3}$[/tex] (since [tex]\(-\frac{3}{2} \cdot \frac{2}{3} = -1\)[/tex]) and the [tex]\(y\)[/tex]-intercept should be 2.
### Answer:
None of the provided options exactly match the necessary criteria of having the perpendicular slope [tex]$\frac{2}{3}$[/tex] and [tex]\(y\)[/tex]-intercept of [tex]$2$[/tex]. Consequently, none of the options given in the question fully meet the criteria for being perpendicular to the given line and having the same [tex]\(y\)[/tex]-intercept.
1. Perpendicular Slopes: Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex]. If one line has a slope [tex]\(m\)[/tex], the perpendicular line must have a slope [tex]\(m_{\perp}\)[/tex] such that [tex]\(m \cdot m_{\perp} = -1\)[/tex].
2. Same [tex]\(y\)[/tex]-Intercept: The [tex]\(y\)[/tex]-intercept of a line is the point where it crosses the [tex]\(y\)[/tex]-axis. For a line to have the same [tex]\(y\)[/tex]-intercept, it must cross the [tex]\(y\)[/tex]-axis at the same point as the given line.
Given:
- The slope of the original line is [tex]\(-\frac{3}{2}\)[/tex] and it contains the point [tex]\((0, 2)\)[/tex], which indicates the [tex]\(y\)[/tex]-intercept is [tex]\(2\)[/tex].
Let's consider each given option to find out which one meets both conditions.
### Option 1: Slope [tex]$-\frac{3}{2}$[/tex], [tex]\(y\)[/tex]-intercept at [tex]$(0, 2)$[/tex].
- Slope = [tex]$-\frac{3}{2}$[/tex].
- The line passes through the point [tex]$(0, 2)$[/tex] so the [tex]\(y\)[/tex]-intercept is [tex]$2$[/tex].
This line has the correct [tex]\(y\)[/tex]-intercept, but the slope is not the negative reciprocal of the given slope [tex]\( -\frac{3}{2} \)[/tex]. Therefore, it is not perpendicular.
### Option 2: Slope [tex]$-\frac{2}{3}$[/tex], [tex]\(y\)[/tex]-intercept at [tex]$(0, -2)$[/tex].
- Slope = [tex]$-\frac{2}{3}$[/tex].
- The line passes through the point [tex]$(0, -2)$[/tex] so the [tex]\(y\)[/tex]-intercept is [tex]$-2$[/tex].
This [tex]\(y\)[/tex]-intercept is different from [tex]$2$[/tex], and the slope also does not satisfy the perpendicular condition. Thus, this option does not meet our criteria.
### Option 3: Slope [tex]$\frac{3}{2}$[/tex], [tex]\(y\)[/tex]-intercept at [tex]$(0, 2)$[/tex].
- Slope = [tex]$\frac{3}{2}$[/tex].
- The line passes through the point [tex]$(0, 2)$[/tex] so the [tex]\(y\)[/tex]-intercept is [tex]$2$[/tex].
This line has the correct [tex]\(y\)[/tex]-intercept, but for perpendicular slopes, the product with the given slope [tex]$\left(-\frac{3}{2}\right) \cdot \frac{3}{2} = -\frac{9}{4}$[/tex], which is not [tex]\(-1\)[/tex]. Hence, it is not perpendicular.
### Option 4: Slope [tex]$-\frac{3}{2}$[/tex], [tex]\(y\)[/tex]-intercept at [tex]$(0, -2)$[/tex].
- Slope = [tex]$-\frac{3}{2}$[/tex].
- The line passes through the point [tex]$(0, -2)$[/tex] so the [tex]\(y\)[/tex]-intercept is [tex]$-2$[/tex].
This [tex]\(y\)[/tex]-intercept is different from [tex]$2$[/tex], and the slope also does not satisfy the perpendicular condition. Thus, this option does not meet our criteria.
### Correct Criteria
According to our criteria:
- The slope of the perpendicular line must be [tex]$0.6666666666666666$[/tex].
- The [tex]\(y\)[/tex]-intercept must be [tex]$2$[/tex].
The slope that satisfies the perpendicular condition to [tex]\(-\frac{3}{2}\)[/tex] is indeed [tex]$\frac{2}{3}$[/tex] (since [tex]\(-\frac{3}{2} \cdot \frac{2}{3} = -1\)[/tex]) and the [tex]\(y\)[/tex]-intercept should be 2.
### Answer:
None of the provided options exactly match the necessary criteria of having the perpendicular slope [tex]$\frac{2}{3}$[/tex] and [tex]\(y\)[/tex]-intercept of [tex]$2$[/tex]. Consequently, none of the options given in the question fully meet the criteria for being perpendicular to the given line and having the same [tex]\(y\)[/tex]-intercept.
