To determine which set of points lies on the line represented by the equation \( y + 2 = \frac{1}{3}(x - 12) \), we can rewrite the equation in slope-intercept form \( y = mx + b \).
Starting with the original equation:
\[ y + 2 = \frac{1}{3}(x - 12) \]
Distributing the right side:
\[ y + 2 = \frac{1}{3}x - 4 \]
Now, subtracting 2 from both sides:
\[ y = \frac{1}{3}x - 6 \]
Now we have our slope-intercept form: \( y = \frac{1}{3}x - 6 \).
Now we can test each set of points to see if they lie on this line.
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Check Point (12, -2): \[ y = \frac{1}{3}(12) - 6 = 4 - 6 = -2 \quad \text{(True)} \] Check Point (15, -1): \[ y = \frac{1}{3}(15) - 6 = 5 - 6 = -1 \quad \text{(True)} \] Both points (12, -2) and (15, -1) lie on the line.
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Check Point (-12, 2): \[ y = \frac{1}{3}(-12) - 6 = -4 - 6 = -10 \quad \text{(False)} \] Check Point (-9, 3): \[ y = \frac{1}{3}(-9) - 6 = -3 - 6 = -9 \quad \text{(False)} \] Neither point (-12, 2) nor (-9, 3) lies on the line.
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Check Point (12, -2): \[ y = \frac{1}{3}(12) - 6 = 4 - 6 = -2 \quad \text{(True)} \] Check Point (13, 1): \[ y = \frac{1}{3}(13) - 6 \approx 4.33 - 6 \approx -1.67 \quad \text{(False)} \] Point (12, -2) lies on the line, but (13, 1) does not.
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Check Point (-12, 2): \[ y = \frac{1}{3}(-12) - 6 = -4 - 6 = -10 \quad \text{(False)} \] Check Point (-11, 3): \[ y = \frac{1}{3}(-11) - 6 \approx -3.67 - 6 \approx -9.67 \quad \text{(False)} \] Neither point (-12, 2) nor (-11, 3) lies on the line.
Conclusion: The only set of points that lies on the line is: (12, -2) and (15, -1).