Find the equation of the line perpendicular to the line \(3x - 2y = 6\) and passes through the point (1, 2).

Step 1 ：Step 1: We first find the slope of the given line. The equation of the line is in the standard form \(Ax + By = C\). We can convert it to the slope-intercept form \(y = mx + b\) by solving for y. \[y = \frac{3x}{2} - 3\] The slope \(m\) of the given line is \(\frac{3}{2}\).

Step 2 ：Step 2: The slope of the line perpendicular to the given line is the negative reciprocal of the slope of the given line. So, the slope \(m'\) of the perpendicular line is \[-\frac{1}{m} = -\frac{2}{3}\].

Step 3 ：Step 3: Now we know the slope of the perpendicular line and a point it passes through (1, 2), we can find the equation of the perpendicular line using the point-slope form \(y - y1 = m'(x - x1)\). Substituting \(m' = -\frac{2}{3}\), \(x1 = 1\), and \(y1 = 2\) into the equation, we get \[y - 2 = -\frac{2}{3}(x - 1)\].

Step 4 ：Step 4: Finally, we simplify the above equation to find the equation of the perpendicular line. \[y = -\frac{2}{3}x + \frac{4}{3} + 2\] \[y = -\frac{2}{3}x + \frac{10}{3}\]