Given the equation of a line as \(y = 3x + 2\), find the equation of the line that is perpendicular to it and passes through the point (2, -1).

Step 1 ：Firstly, we know that the slope of two perpendicular lines are negative reciprocals. Hence, the slope of the line we are looking for is \(-1/3\), since it's the negative reciprocal of the slope of the given line which is 3.

Step 2 ：Secondly, we use the point-slope form of the equation of a line, which is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope of the line. Here, \(x_1 = 2\), \(y_1 = -1\) and \(m = -1/3\). Substituting these values in, we get the equation \(y - (-1) = -1/3 (x - 2)\).

Step 3 ：Finally, simplifying the equation, we get \(y + 1 = -1/3x + 2/3\), which rearranges to \(y = -1/3x + 2/3 - 1\).