To determine whether the given pairs of equations represent parallel, perpendicular, or neither lines, we need to analyze their slopes. Here's how:

1. Convert each equation to slope-intercept form $y = mx + b$, where $m$ is the slope.
2. If two lines have the same slope ($m_1 = m_2$), they are parallel.
3. If two lines have slopes that are negative reciprocals of each other ($m_1 = -\frac{1}{m_2}$), they are perpendicular.
4. If neither condition is met, the lines are neither.

Let's go through each pair.

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(a) $6x - 12y = 24$ and $4x + 2y = 8$

First equation: $6x - 12y = 24$

• Solve for $y$: $6x - 12y = 24 \quad \Rightarrow \quad -12y = -6x + 24 \quad \Rightarrow \quad y = \frac{1}{2}x - 2$ The slope is $m_1 = \frac{1}{2}$.

Second equation: $4x + 2y = 8$

• Solve for $y$: $2y = -4x + 8 \quad \Rightarrow \quad y = -2x + 4$ The slope is $m_2 = -2$.

• The slopes $m_1 = \frac{1}{2}$ and $m_2 = -2$ are negative reciprocals of each other, meaning the lines are perpendicular.

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(b) $4x + y = 5$ and $3x + 12y = -6$

First equation: $4x + y = 5$

• Solve for $y$: $y = -4x + 5$ The slope is $m_1 = -4$.

Second equation: $3x + 12y = -6$

• Solve for $y$: $12y = -3x - 6 \quad \Rightarrow \quad y = -\frac{1}{4}x - \frac{1}{2}$ The slope is $m_2 = -\frac{1}{4}$.

• The slopes $m_1 = -4$ and $m_2 = -\frac{1}{4}$ are neither equal nor negative reciprocals, so the lines are neither parallel nor perpendicular.

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(c) $-2x + 7y = 14$ and $4x = 14y$

First equation: $-2x + 7y = 14$

• Solve for $y$: $7y = 2x + 14 \quad \Rightarrow \quad y = \frac{2}{7}x + 2$ The slope is $m_1 = \frac{2}{7}$.

Second equation: $4x = 14y$

• Solve for $y$: $y = \frac{2}{7}x$ The slope