To find the equation of the line that passes through the point (1, –2) and is parallel to the line given by the equation \( y = 3x - 2 \), we can follow these steps:
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Identify the Slope: The slope of the line \( y = 3x - 2 \) is 3 (since it is in the form \( y = mx + b \), where \( m \) is the slope). Since parallel lines have the same slope, the slope of our new line will also be 3.
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Use the Point-Slope Form: We can use the point-slope form of the equation of a line, which is given by: \[ y - y_1 = m(x - x_1) \] where \( (x_1, y_1) \) is the point the line passes through and \( m \) is the slope. Here, \( (x_1, y_1) = (1, -2) \) and \( m = 3 \).
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Substitute the Values: Plugging the values into the point-slope form: \[ y - (-2) = 3(x - 1) \] Simplifying this, we get: \[ y + 2 = 3(x - 1) \]
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Distribute the Slope: \[ y + 2 = 3x - 3 \]
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Isolate y: Now, we need to solve for \( y \): \[ y = 3x - 3 - 2 \] \[ y = 3x - 5 \]
Thus, the equation of the line in slope-intercept form that passes through the point (1, –2) and is parallel to the line \( y = 3x - 2 \) is: \[ \boxed{y = 3x - 5} \]
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