Answer :
Sure, let's work through each part of the question step-by-step.
### Part 1: Identifying which line is neither parallel nor perpendicular to [tex]\(x + 2y = 1\)[/tex]
1. Find the slope of the given line [tex]\(x + 2y = 1\)[/tex]:
- Rearrange this equation into slope-intercept form [tex]\(y = mx + b\)[/tex].
[tex]\[ x + 2y = 1 \\ 2y = -x + 1 \\ y = -\frac{1}{2}x + \frac{1}{2} \quad \text{(Slope \(m\) is \(-\frac{1}{2}\))} \][/tex]
2. Identify the slopes of the other lines:
- Line [tex]\(A: y = 2x - 1\)[/tex]:
[tex]\[ \text{The slope (\(m_A\)) is } 2 \][/tex]
- Line [tex]\(B: x = 2y - 1\)[/tex]:
- Rearrange into slope-intercept form [tex]\(y = mx + b\)[/tex]:
[tex]\[ x = 2y - 1 \\ 2y = x + 1 \\ y = \frac{1}{2}x + \frac{1}{2} \quad \text{(Slope \(m_B\) is \(\frac{1}{2}\))} \][/tex]
- Line [tex]\(C: y = -\frac{1}{2}x + 3\)[/tex]:
[tex]\[ \text{The slope (\(m_C\)) is } -\frac{1}{2} \][/tex]
3. Determine which lines are parallel or perpendicular to the given line [tex]\(x + 2y = 1\)[/tex]:
- Lines are parallel if their slopes are equal.
- Lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex].
4. Evaluate relationships:
- Given line's slope: [tex]\(-\frac{1}{2}\)[/tex].
- Line [tex]\(A\)[/tex] with slope [tex]\(2\)[/tex]:
[tex]\[ \text{Not parallel and not perpendicular since } 2 \times (-\frac{1}{2}) = -1 \quad \text{(perpendicular)} \][/tex]
- Line [tex]\(B\)[/tex] with slope [tex]\( \frac{1}{2} \)[/tex]:
[tex]\[ \text{Not parallel and not perpendicular since } \frac{1}{2} \neq -\frac{1}{2} \quad \text{(not perpendicular, and not equal to \(-\frac{1}{2}\))} \][/tex]
- Line [tex]\(C\)[/tex] with slope [tex]\(-\frac{1}{2}\)[/tex]:
[tex]\[ \text{Parallel because } -\frac{1}{2} = -\frac{1}{2} \][/tex]
Therefore, the line from options [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] which is neither parallel nor perpendicular to the given line [tex]\(x+2y=1\)[/tex] is:
[tex]\[ \boxed{B: x = 2y - 1} \][/tex]
### Part 2: Finding values of [tex]\(a\)[/tex] for lines to be parallel
1. Given lines [tex]\(y = ax - 3\)[/tex] and [tex]\(y = \frac{3}{a}x\)[/tex]:
- For these lines to be parallel, their slopes must be equal:.
[tex]\[ \text{Slopes of } y = ax - 3 \quad \text{and} \quad y = \frac{3}{a}x \\ \text{are } a \text{ and } \frac{3}{a} \text{ respectively. They must be equal:} \][/tex]
[tex]\[ a = \frac{3}{a} \][/tex]
2. Solve for [tex]\(a\)[/tex]:
- Multiply both sides by [tex]\(a\)[/tex]:
[tex]\[ a^2 = 3 \][/tex]
- Take the square root of both sides:
[tex]\[ a = \sqrt{3} \quad \text{or} \quad a = -\sqrt{3} \][/tex]
Therefore, the set of values of [tex]\(a\)[/tex] such that the two lines are parallel is:
[tex]\[ \boxed{\left\{\sqrt{3}, -\sqrt{3}\right\}} \][/tex]
### Part 1: Identifying which line is neither parallel nor perpendicular to [tex]\(x + 2y = 1\)[/tex]
1. Find the slope of the given line [tex]\(x + 2y = 1\)[/tex]:
- Rearrange this equation into slope-intercept form [tex]\(y = mx + b\)[/tex].
[tex]\[ x + 2y = 1 \\ 2y = -x + 1 \\ y = -\frac{1}{2}x + \frac{1}{2} \quad \text{(Slope \(m\) is \(-\frac{1}{2}\))} \][/tex]
2. Identify the slopes of the other lines:
- Line [tex]\(A: y = 2x - 1\)[/tex]:
[tex]\[ \text{The slope (\(m_A\)) is } 2 \][/tex]
- Line [tex]\(B: x = 2y - 1\)[/tex]:
- Rearrange into slope-intercept form [tex]\(y = mx + b\)[/tex]:
[tex]\[ x = 2y - 1 \\ 2y = x + 1 \\ y = \frac{1}{2}x + \frac{1}{2} \quad \text{(Slope \(m_B\) is \(\frac{1}{2}\))} \][/tex]
- Line [tex]\(C: y = -\frac{1}{2}x + 3\)[/tex]:
[tex]\[ \text{The slope (\(m_C\)) is } -\frac{1}{2} \][/tex]
3. Determine which lines are parallel or perpendicular to the given line [tex]\(x + 2y = 1\)[/tex]:
- Lines are parallel if their slopes are equal.
- Lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex].
4. Evaluate relationships:
- Given line's slope: [tex]\(-\frac{1}{2}\)[/tex].
- Line [tex]\(A\)[/tex] with slope [tex]\(2\)[/tex]:
[tex]\[ \text{Not parallel and not perpendicular since } 2 \times (-\frac{1}{2}) = -1 \quad \text{(perpendicular)} \][/tex]
- Line [tex]\(B\)[/tex] with slope [tex]\( \frac{1}{2} \)[/tex]:
[tex]\[ \text{Not parallel and not perpendicular since } \frac{1}{2} \neq -\frac{1}{2} \quad \text{(not perpendicular, and not equal to \(-\frac{1}{2}\))} \][/tex]
- Line [tex]\(C\)[/tex] with slope [tex]\(-\frac{1}{2}\)[/tex]:
[tex]\[ \text{Parallel because } -\frac{1}{2} = -\frac{1}{2} \][/tex]
Therefore, the line from options [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] which is neither parallel nor perpendicular to the given line [tex]\(x+2y=1\)[/tex] is:
[tex]\[ \boxed{B: x = 2y - 1} \][/tex]
### Part 2: Finding values of [tex]\(a\)[/tex] for lines to be parallel
1. Given lines [tex]\(y = ax - 3\)[/tex] and [tex]\(y = \frac{3}{a}x\)[/tex]:
- For these lines to be parallel, their slopes must be equal:.
[tex]\[ \text{Slopes of } y = ax - 3 \quad \text{and} \quad y = \frac{3}{a}x \\ \text{are } a \text{ and } \frac{3}{a} \text{ respectively. They must be equal:} \][/tex]
[tex]\[ a = \frac{3}{a} \][/tex]
2. Solve for [tex]\(a\)[/tex]:
- Multiply both sides by [tex]\(a\)[/tex]:
[tex]\[ a^2 = 3 \][/tex]
- Take the square root of both sides:
[tex]\[ a = \sqrt{3} \quad \text{or} \quad a = -\sqrt{3} \][/tex]
Therefore, the set of values of [tex]\(a\)[/tex] such that the two lines are parallel is:
[tex]\[ \boxed{\left\{\sqrt{3}, -\sqrt{3}\right\}} \][/tex]
