To determine if pairs of lines are parallel, perpendicular, or intersecting, we need to find their slopes.
1st Pair of Lines
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Line 1: \( y = -\frac{2}{3}x + 7 \)
Slope (m1) = \(-\frac{2}{3}\) -
Line 2: Rearranging \( -2x - 3y = 12 \) to slope-intercept form: \[ -3y = 2x + 12 \implies y = -\frac{2}{3}x - 4 \] Slope (m2) = \(-\frac{2}{3}\)
Since the slopes are equal (\(m1 = m2\)), the lines are parallel.
2nd Pair of Lines
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Line 1: Rearranging \( 2x - y = 7 \) to slope-intercept form: \[ -y = -2x + 7 \implies y = 2x - 7 \] Slope (m1) = \(2\)
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Line 2: Rearranging \( 3x - 4y = -8 \) to slope-intercept form: \[ -4y = -3x - 8 \implies y = \frac{3}{4}x + 2 \] Slope (m2) = \(\frac{3}{4}\)
Since the slopes are different, we check if they are perpendicular:
- Two lines are perpendicular if the product of their slopes is \(-1\): \[ 2 \times \frac{3}{4} = \frac{3}{2} \quad (\text{not }-1) \]
Thus, the lines are intersecting.
3rd Pair of Lines
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Line 1: Rearranging \(y - 6 = \frac{1}{4}(x + 2)\): \[ y - 6 = \frac{1}{4}x + \frac{1}{2} \implies y = \frac{1}{4}x + 6.5 \] Slope (m1) = \(\frac{1}{4}\)
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Line 2: The equation is already in slope-intercept form \(y = -4x - 10\): Slope (m2) = \(-4\)
Since the slopes are different, we again check if they are perpendicular: \[ \frac{1}{4} \times (-4) = -1 \] Since the product of the slopes is \(-1\), the lines are perpendicular.
Summary:
- First pair: Parallel
- Second pair: Intersecting
- Third pair: Perpendicular