To determine which two lines are perpendicular, we need to find the slopes of each line and check if the product of their slopes equals -1 (the condition for perpendicular lines).
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Line 1: \(5x - 3y = -3\)
Rearranging to slope-intercept form (y = mx + b): \(-3y = -5x - 3\) \(y = \frac{5}{3}x + 1\)
Slope (m) = \(\frac{5}{3}\)
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Line 2: \(4x - 3y = -6\)
Rearranging to slope-intercept form: \(-3y = -4x - 6\) \(y = \frac{4}{3}x + 2\)
Slope (m) = \(\frac{4}{3}\)
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Line 3: \(y - 2 = \frac{3}{4}(x + 1)\)
This is already in point-slope form. Rearranging to slope-intercept: \(y - 2 = \frac{3}{4}x + \frac{3}{4}\) \(y = \frac{3}{4}x + \frac{11}{4}\)
Slope (m) = \(\frac{3}{4}\)
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Line 4: \(y = -\frac{3}{4}x + 12\)
This is already in slope-intercept form.
Slope (m) = \(-\frac{3}{4}\)
Now we can compare the slopes:
- Slope of Line 1 = \(\frac{5}{3}\)
- Slope of Line 2 = \(\frac{4}{3}\)
- Slope of Line 3 = \(\frac{3}{4}\)
- Slope of Line 4 = \(-\frac{3}{4}\)
Product of slopes for perpendicularity:
The lines with slopes \(\frac{3}{4}\) and \(-\frac{3}{4}\) multiply to: \[ \frac{3}{4} \cdot -\frac{3}{4} = -\frac{9}{16} \] which does not equal -1.
The lines with slopes \(\frac{5}{3}\) and \(-\frac{3}{4}\) multiply to: \[ \frac{5}{3} \cdot -\frac{3}{4} = -\frac{15}{12} = -\frac{5}{4} \] which also does not equal -1.
However, if we take the negative inverse of \(\frac{3}{4}\), which is \(-\frac{4}{3}\), we find none of the other pairs hold true either.
Thus, the pair of lines that are perpendicular is:
y = -3/4 x + 12 and y - 2 = 3/4 (x + 1)
These two lines (the last two) have slopes that are negative reciprocals to each other:
- \(y = \frac{3}{4}x + b_3\) has a slope of \(\frac{3}{4}\).
- \(y = -\frac{3}{4}x + b_4\) has a slope of \(-\frac{3}{4}\).
To clarify, the final selection is:
- Line 3: \(y - 2 = \frac{3}{4}(x + 1)\)
- Line 4: \(y = -\frac{3}{4}x + 12\)
They are perpendicular.