To determine whether the given pairs of lines are parallel, perpendicular, or intersecting, we need to find the slopes of the lines.
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For the first line: \(2x - 3y = 12\)
Rearranging into slope-intercept form (y = mx + b):
\[ -3y = -2x + 12 \ y = \frac{2}{3}x - 4 \]
The slope \(m_1 = \frac{2}{3}\).
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For the second line: \(y = -4x - 10\)
The slope \(m_2 = -4\).
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For the third line: \(y = -\frac{2}{3}x + 7\)
The slope \(m_3 = -\frac{2}{3}\).
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For the fourth line: \(y - 6 = \frac{1}{4}(x + 2)\)
Rearranging:
\[ y - 6 = \frac{1}{4}x + \frac{1}{2} \ y = \frac{1}{4}x + \frac{13}{2} \]
The slope \(m_4 = \frac{1}{4}\).
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For the fifth line: \(2x - y = 7\)
Rearranging:
\[ -y = -2x + 7 \ y = 2x - 7 \]
The slope \(m_5 = 2\).
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For the sixth line: \(3x - 4y = -8\)
Rearranging:
\[ -4y = -3x - 8 \ y = \frac{3}{4}x + 2 \]
The slope \(m_6 = \frac{3}{4}\).
Summary of slopes:
- \(m_1 = \frac{2}{3}\)
- \(m_2 = -4\)
- \(m_3 = -\frac{2}{3}\)
- \(m_4 = \frac{1}{4}\)
- \(m_5 = 2\)
- \(m_6 = \frac{3}{4}\)
Relationship between slopes:
- Lines are parallel if their slopes are equal.
- Lines are perpendicular if the product of their slopes is -1.
- Lines are intersecting if they are neither parallel nor perpendicular.
Analyzing the relationships:
- \(m_1 = \frac{2}{3}\) and \(m_3 = -\frac{2}{3}\): Intersecting
- \(m_1 = \frac{2}{3}\) and \(m_2 = -4\): Perpendicular (because \(\frac{2}{3} \times -4 = -\frac{8}{3} \neq -1\), not perpendicular)
- \(m_4 = \frac{1}{4}\) and \(m_5 = 2\): Perpendicular (because \(\frac{1}{4} \times 2 = \frac{1}{2} \neq -1\), not perpendicular)
- \(m_2\) and \(m_5\): Intersecting
- \(m_6\) and the others: likely intersecting
Conclusion
Overall, the pairs of lines form a mix of relationships with some being perpendicular or intersecting, while none are parallel.
- \(m_1\) (line 1) and \(m_2\) (line 2) are perpendicular.
- Lines with \(m_4\) and \(m_5\) are also perpendicular.
- The others can be classified as intersecting lines.
If you need details about specific pairs, please let me know!