To find the equation of the line that is perpendicular to the given line and passes through the point (2, -1), we first need to rearrange the equation of the given line \(3x - 2y = 7\) into slope-intercept form \(y = mx + b\).
- Rearrange the equation: \[ 3x - 2y = 7 \] Solve for \(y\): \[ -2y = -3x + 7 \] Dividing by -2: \[ y = \frac{3}{2}x - \frac{7}{2} \]
The slope \(m\) of the given line is \(\frac{3}{2}\).
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The slope of a line that is perpendicular to another line is the negative reciprocal of the original line's slope. Therefore, the slope of the line we want to find is: \[ m_{\text{perpendicular}} = -\frac{1}{m} = -\frac{1}{\frac{3}{2}} = -\frac{2}{3} \]
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Now we have the slope of the perpendicular line, which is \(-\frac{2}{3}\), and it passes through the point (2, -1). We can use the point-slope form of the equation of a line: \[ y - y_1 = m(x - x_1) \] Where \((x_1, y_1)\) is the point (2, -1).
Substituting the values: \[ y - (-1) = -\frac{2}{3}(x - 2) \] This simplifies to: \[ y + 1 = -\frac{2}{3}(x - 2) \]
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Now, let's simplify this equation: Distributing \(-\frac{2}{3}\): \[ y + 1 = -\frac{2}{3}x + \frac{4}{3} \] Subtracting 1 from both sides: \[ y = -\frac{2}{3}x + \frac{4}{3} - 1 \] To combine the constant terms, rewrite 1 as \(\frac{3}{3}\): \[ y = -\frac{2}{3}x + \frac{4}{3} - \frac{3}{3} \] \[ y = -\frac{2}{3}x + \frac{1}{3} \]
Thus, the equation of the line that is perpendicular to \(3x - 2y = 7\) and passes through the point (2, -1) is: \[ y = -\frac{2}{3}x + \frac{1}{3} \]