To find the equation of the line passing through the point (2, -2) and perpendicular to the given line, we first need to determine the slope of the given line.
The equation \(3x - 2y = 7\) can be rearranged into slope-intercept form \(y = mx + b\), where \(m\) is the slope.
- Rearranging the equation: \[ -2y = -3x + 7 \] \[ y = \frac{3}{2}x - \frac{7}{2} \]
From this, we can see that the slope \(m\) of the given line is \(\frac{3}{2}\).
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The slope of the line that is perpendicular to this line is the negative reciprocal of \(\frac{3}{2}\). Thus, the slope \(m'\) of the perpendicular line is: \[ m' = -\frac{1}{\left(\frac{3}{2}\right)} = -\frac{2}{3} \]
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Now we have the slope of the perpendicular line and the point (2, -2) through which it passes. We can use the point-slope form of the line equation: \[ y - y_1 = m'(x - x_1) \] Substituting in the point (2, -2) and the slope \(m' = -\frac{2}{3}\): \[ y - (-2) = -\frac{2}{3}(x - 2) \] \[ y + 2 = -\frac{2}{3}(x - 2) \] \[ y + 2 = -\frac{2}{3}x + \frac{4}{3} \]
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Now, isolate \(y\): \[ y = -\frac{2}{3}x + \frac{4}{3} - 2 \] To combine the constants: \[ -2 = -\frac{6}{3}, \quad \text{therefore} \] \[ y = -\frac{2}{3}x + \frac{4}{3} - \frac{6}{3} \] \[ y = -\frac{2}{3}x - \frac{2}{3} \]
Thus, the equation of the line passing through the point (2, -2) and perpendicular to the line \(3x - 2y = 7\) is: \[ y = -\frac{2}{3}x - \frac{2}{3} \]