To find the equation of a line that is perpendicular to the given line \(4x - 12y = 2\) and passes through the point \((10, -1)\), we need to follow these steps:
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Rewrite the given line in slope-intercept form (y = mx + b): \[ 4x - 12y = 2 \] First, isolate \(y\): \[ -12y = -4x + 2 \] Then, divide by -12: \[ y = \frac{4}{12}x - \frac{2}{12} \] Simplifying gives: \[ y = \frac{1}{3}x - \frac{1}{6} \]
The slope \(m_1\) of the original line is \(\frac{1}{3}\).
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Find the slope of the perpendicular line: The slope of a line perpendicular to another is the negative reciprocal of the original line's slope. Thus, \[ m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{1}{3}} = -3 \]
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Use the point-slope form to write the equation of the perpendicular line: The point-slope form of a line is given by: \[ y - y_1 = m(x - x_1) \] Plugging in the point \((10, -1)\) and the new slope \(-3\): \[ y - (-1) = -3(x - 10) \] Simplifying this: \[ y + 1 = -3(x - 10) \] \[ y + 1 = -3x + 30 \] \[ y = -3x + 30 - 1 \] \[ y = -3x + 29 \]
Thus, the equation of the line that is perpendicular to \(4x - 12y = 2\) and passes through the point \((10, -1)\) is: \[ \boxed{y = -3x + 29} \]