To determine the relationship between the pairs of lines given by their equations, we can convert all of the equations to slope-intercept form (y = mx + b), where m represents the slope. Two lines are parallel if they have the same slope, perpendicular if the product of their slopes is -1, and intersecting if neither condition applies.
Let's find the slopes for each pair of lines.
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Different pairs of lines:
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First Pair:
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\( 2x - y = 7 \) \[ y = 2x - 7 \quad \text{(slope } m_1 = 2\text{)} \]
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\( 3x - 4y = -8 \) \[ 4y = 3x + 8 \implies y = \frac{3}{4}x + 2 \quad \text{(slope } m_2 = \frac{3}{4}\text{)} \]
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Since \( m_1 \neq m_2 \) and \( m_1 \cdot m_2 \neq -1 \), these lines are intersecting.
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Second Pair:
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\( y - 6 = \frac{1}{4}(x + 2) \) \[ y - 6 = \frac{1}{4}x + \frac{1}{2} \implies y = \frac{1}{4}x + 6.5 \quad \text{(slope } m_3 = \frac{1}{4}\text{)} \]
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\( y = -4x - 10 \) \[ (slope } m_4 = -4\text{)} \]
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Since \( m_3 \cdot m_4 = \frac{1}{4} \cdot -4 = -1 \), these lines are perpendicular.
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Third Pair:
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\( y = -\frac{2}{3}x + 7 \) \[ (slope } m_5 = -\frac{2}{3}\text{)} \]
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\( -2x - 3y = 12 \) \[ 3y = -2x - 12 \implies y = -\frac{2}{3}x - 4 \quad \text{(slope } m_6 = -\frac{2}{3}\text{)} \]
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Since \( m_5 = m_6 \), these lines are parallel.
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In summary:
- The first pair is intersecting.
- The second pair is perpendicular.
- The third pair is parallel.