Find the equation of the line perpendicular to the line \(3x - 4y = 8\) and passes through the point \((2, -3)\).

Step 1 ：Step 1: Find the slope of the given line. The slope-intercept form of a line equation is \(y = mx + c\), where m is the slope. Let's rearrange \(3x - 4y = 8\) to this form: \(y = \frac{3}{4}x - 2\). So, the slope (m1) of the given line is \(\frac{3}{4}\).

Step 2 ：Step 2: The slope of a line perpendicular to a line with slope m1 is \(-\frac{1}{m1}\). Therefore, the slope (m2) of the line perpendicular to the given line is \(-\frac{1}{\frac{3}{4}} = -\frac{4}{3}\).

Step 3 ：Step 3: Now, we know the slope of the line we want to find and a point \((2, -3)\) it passes through. We can use the point-slope form of the line equation, which is \(y - y1 = m2(x - x1)\), where \((x1, y1)\) is the known point. Substituting the known values, we get \(y - (-3) = -\frac{4}{3}(x - 2)\), which simplifies to \(y + 3 = -\frac{4}{3}x + \frac{8}{3}\).

Step 4 ：Step 4: Finally, let's rearrange this to the slope-intercept form: \(y = -\frac{4}{3}x + \frac{8}{3} - 3\), which simplifies to \(y = -\frac{4}{3}x - \frac{1}{3}\).