To find the equation of a line that is perpendicular to the line given by the equation \(2y = 3x - 10\) and passes through the point \((-6, 1)\), follow these steps:
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Rearrange the given equation into slope-intercept form (y = mx + b): \[ 2y = 3x - 10 \] Divide both sides by 2: \[ y = \frac{3}{2}x - 5 \] Here, the slope \(m\) of this line is \(\frac{3}{2}\).
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Find the slope of the perpendicular line: The slope of any line perpendicular to another line is the negative reciprocal of the original line's slope. Therefore, the slope of the line we want is: \[ m_{\text{perpendicular}} = -\frac{1}{\frac{3}{2}} = -\frac{2}{3} \]
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Use point-slope form of the equation of a line: The point-slope form is given by: \[ y - y_1 = m(x - x_1) \] where \((x_1, y_1)\) is the point \((-6, 1)\) and \(m = -\frac{2}{3}\).
Plugging in the values: \[ y - 1 = -\frac{2}{3}(x + 6) \]
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Simplify the equation: Distributing the slope: \[ y - 1 = -\frac{2}{3}x - 4 \] Adding 1 to both sides: \[ y = -\frac{2}{3}x - 3 \]
So, the equation of the line that is perpendicular to \(2y = 3x - 10\) and passes through the point \((-6, 1)\) is: \[ \boxed{y = -\frac{2}{3}x - 3} \]