To determine which set of points lies on the line represented by the equation \( y + 2 = \frac{1}{3}(x - 12) \), we first need to convert this equation into a more standard slope-intercept form, \( y = mx + b \).
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Start with the given equation: \[ y + 2 = \frac{1}{3}(x - 12) \]
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Distributing the \(\frac{1}{3}\): \[ y + 2 = \frac{1}{3}x - 4 \]
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Now isolate \(y\): \[ y = \frac{1}{3}x - 4 - 2 \] \[ y = \frac{1}{3}x - 6 \]
Now that we have the equation of the line in slope-intercept form (\(y = \frac{1}{3}x - 6\)), we can substitute the x-coordinates of each given point to see if the corresponding y-values match.
Check each option:
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Option (−12, 2) and (−9, 3):
- For \(x = -12\): \[ y = \frac{1}{3}(-12) - 6 = -4 - 6 = -10 \quad (\text{not } 2) \]
- For \(x = -9\): \[ y = \frac{1}{3}(-9) - 6 = -3 - 6 = -9 \quad (\text{not } 3) \]
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Option (12, -2) and (13, 1):
- For \(x = 12\): \[ y = \frac{1}{3}(12) - 6 = 4 - 6 = -2 \quad (\text{correct!}) \]
- For \(x = 13\): \[ y = \frac{1}{3}(13) - 6 \approx 4.33 - 6 = -1.67 \quad (\text{not } 1) \]
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Option (12, -2) and (15, -1):
- For \(x = 12\): \[ y = \frac{1}{3}(12) - 6 = 4 - 6 = -2 \quad (\text{correct!}) \]
- For \(x = 15\): \[ y = \frac{1}{3}(15) - 6 = 5 - 6 = -1 \quad (\text{correct!}) \]
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Option (−12, 2) and (−11, 3):
- For \(x = -12\): \[ y = \frac{1}{3}(-12) - 6 = -10 \quad (\text{not } 2) \]
- For \(x = -11\): \[ y = \frac{1}{3}(-11) - 6 \approx -3.67 - 6 = -9.67 \quad (\text{not } 3) \]
Conclusion:
The only correct option where both points lie on the line is (12, −2) and (15, −1).