To solve the problem, we need to determine how each of the given equations compares to the line [tex]\(a\)[/tex], which is represented by the equation [tex]\(y = -2x + 3\)[/tex].
### Step-by-Step Solution:
1. Identify the slope of line [tex]\(a\)[/tex]:
The slope-intercept form of a line is [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope.
For the line [tex]\(a\)[/tex], [tex]\(y = -2x + 3\)[/tex], the slope [tex]\(m\)[/tex] is [tex]\(-2\)[/tex].
2. Compare the slopes of the given equations to the slope of line [tex]\(a\)[/tex]:
- Equation 1: [tex]\(y = 2x - 1\)[/tex]
- The slope is [tex]\(2\)[/tex].
- Equation 2: [tex]\(y = -2x + 5\)[/tex]
- The slope is [tex]\(-2\)[/tex].
- Equation 3: [tex]\(y = \frac{1}{2}x + 7\)[/tex]
- The slope is [tex]\(\frac{1}{2}\)[/tex].
3. Determine the relationship of each equation to line [tex]\(a\)[/tex]:
- Parallel Lines:
- Two lines are parallel if they have the same slope.
- The slope of line [tex]\(a\)[/tex] is [tex]\(-2\)[/tex].
- The equation that has the same slope is [tex]\(y = -2x + 5\)[/tex].
- Therefore, [tex]\(y = -2x + 5\)[/tex] is parallel to line [tex]\(a\)[/tex].
- Perpendicular Lines:
- Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex]. This means their slopes are negative reciprocals of each other.
- The slope of line [tex]\(a\)[/tex] is [tex]\(-2\)[/tex].
- The negative reciprocal of [tex]\(-2\)[/tex] is [tex]\(\frac{1}{2}\)[/tex].
- The equation that has the slope [tex]\(\frac{1}{2}\)[/tex] is [tex]\(y = \frac{1}{2}x + 7\)[/tex].
- Therefore, [tex]\(y = \frac{1}{2}x + 7\)[/tex] is perpendicular to line [tex]\(a\)[/tex].
- Neither Parallel Nor Perpendicular:
- For an equation to be neither parallel nor perpendicular, its slope should not be the same as or the negative reciprocal of the slope of line [tex]\(a\)[/tex].
- The slope [tex]\(2\)[/tex] (from [tex]\(y = 2x - 1\)[/tex]) is neither [tex]\(-2\)[/tex] nor [tex]\(\frac{1}{2}\)[/tex].
- Therefore, [tex]\(y = 2x - 1\)[/tex] is neither parallel nor perpendicular to line [tex]\(a\)[/tex].
### Final Conclusion:
[tex]\[
\begin{tabular}{|c|c|c|}
\hline
\text{Parallel to line \(a\)} & \text{Perpendicular to line \(a\)} & \text{Neither parallel nor perpendicular to line \(a\)} \\
\hline
y = -2x + 5 & y = \frac{1}{2}x + 7 & y = 2x - 1 \\
\hline
\end{tabular}
\][/tex]
