Find the equation of the line perpendicular to the line \(3x - 4y = 12\) and passes through the point \((2, -1)\).

Step 1 ：First, we convert the given equation into slope-intercept form (\(y = mx + b\)), where \(m\) is the slope and \(b\) is the y-intercept. So, \(3x - 4y = 12\) becomes \(y = \frac{3}{4}x - 3\). Thus, the slope of the given line is \(\frac{3}{4}\).

Step 2 ：The slope of a line perpendicular to a line with slope \(m\) is \(-\frac{1}{m}\). Therefore, the slope of the line we want to find is \(-\frac{1}{\frac{3}{4}} = -\frac{4}{3}\).

Step 3 ：Now, we use the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line. Substituting the point \((2, -1)\) and the slope \(-\frac{4}{3}\) into the equation, we get \(y - (-1) = -\frac{4}{3}(x - 2)\), which simplifies to \(y + 1 = -\frac{4}{3}x + \frac{8}{3}\).

Step 4 ：Finally, rearranging the equation to slope-intercept form, we get \(y = -\frac{4}{3}x + \frac{8}{3} - 1 = -\frac{4}{3}x + \frac{5}{3}\).