To find the equation of a line that is perpendicular to the given line \( 2x + 9y = -30 \) and passes through the point \((-8, 0)\), we can follow these steps:
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Find the slope of the given line:
We need to rewrite the equation of the line in slope-intercept form (y = mx + b), where m is the slope.
Start with the equation: \[ 2x + 9y = -30 \]
Isolate \(y\): \[ 9y = -2x - 30 \] \[ y = -\frac{2}{9}x - \frac{30}{9} \] \[ y = -\frac{2}{9}x - \frac{10}{3} \]
The slope of the given line \( m_1 = -\frac{2}{9} \).
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Determine the slope of the perpendicular line:
The slope of the line that is perpendicular to another line is the negative reciprocal of the original slope. Thus, \[ m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{2}{9}} = \frac{9}{2} \]
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Use the point-slope formula:
Now that we have the slope \( m_2 = \frac{9}{2} \) and the point \((-8, 0)\), we can use the point-slope form of a line: \[ y - y_1 = m(x - x_1) \]
Plugging in our values: \[ y - 0 = \frac{9}{2}(x - (-8)) \] \[ y = \frac{9}{2}(x + 8) \]
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Simplify the equation:
Distributing the slope: \[ y = \frac{9}{2}x + \frac{9 \cdot 8}{2} \] \[ y = \frac{9}{2}x + 36 \]
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Convert to standard form (if needed):
If we want to express it in standard form \(Ax + By = C\), we can multiply through by 2 to eliminate the fraction: \[ 2y = 9x + 72 \] Rearranging gives us: \[ -9x + 2y = 72 \] or equivalently, \[ 9x - 2y = -72 \]
Thus, the equation of the line that is perpendicular to the given line and passes through the point \((-8, 0)\) is: \[ \boxed{9x - 2y = -72} \]
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