To determine which lines are perpendicular from the given equations, we need to find the slopes of each line. Two lines are perpendicular if the product of their slopes is -1.
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Line 1: \( y - 2 = \frac{3}{4}(x + 1) \)
Rearranging this into slope-intercept form (y = mx + b): \[ y - 2 = \frac{3}{4}x + \frac{3}{4} \] \[ y = \frac{3}{4}x + \frac{3}{4} + 2 \] \[ y = \frac{3}{4}x + \frac{11}{4} \] The slope is \( m_1 = \frac{3}{4} \).
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Line 2: \( y = -\frac{3}{4}x + 12 \)
The slope is \( m_2 = -\frac{3}{4} \).
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Line 3: \( 4x - 3y = -6 \)
Rearranging into slope-intercept form: \[ -3y = -4x - 6 \] \[ y = \frac{4}{3}x + 2 \] The slope is \( m_3 = \frac{4}{3} \).
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Line 4: \( 5x - 3y = -3 \)
Rearranging into slope-intercept form: \[ -3y = -5x - 3 \] \[ y = \frac{5}{3}x + 1 \] The slope is \( m_4 = \frac{5}{3} \).
Now let's check for perpendicularity:
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Line 1: \( m_1 = \frac{3}{4} \)
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Line 2: \( m_2 = -\frac{3}{4} \)
- Product: \( \frac{3}{4} \cdot -\frac{3}{4} = -\frac{9}{16} \) (NOT perpendicular)
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Line 1: \( m_1 = \frac{3}{4} \)
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Line 3: \( m_3 = \frac{4}{3} \)
- Product: \( \frac{3}{4} \cdot \frac{4}{3} = 1 \) (NOT perpendicular)
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Line 1: \( m_1 = \frac{3}{4} \)
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Line 4: \( m_4 = \frac{5}{3} \)
- Product: \( \frac{3}{4} \cdot \frac{5}{3} = \frac{15}{12} \) (NOT perpendicular)
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Line 2: \( m_2 = -\frac{3}{4} \)
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Line 3: \( m_3 = \frac{4}{3} \)
- Product: \( -\frac{3}{4} \cdot \frac{4}{3} = -1 \) (PERPENDICULAR)
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Line 2: \( m_2 = -\frac{3}{4} \)
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Line 4: \( m_4 = \frac{5}{3} \)
- Product: \( -\frac{3}{4} \cdot \frac{5}{3} = -\frac{15}{12} \) (NOT perpendicular)
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Line 3: \( m_3 = \frac{4}{3} \)
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Line 4: \( m_4 = \frac{5}{3} \)
- Product: \( \frac{4}{3} \cdot \frac{5}{3} = \frac{20}{9} \) (NOT perpendicular)
Conclusion: The only pair of lines that are perpendicular is:
- Line 2: \( y = -\frac{3}{4}x + 12 \)
- Line 3: \( 4x - 3y = -6 \)
Thus, the lines that are perpendicular are:
- y = -3/4 x + 12
- 4x - 3y = -6