What is the equation of the line parallel to the line \(3x-4y+7=0\) and passing through the point (2, -3)?

Step 1 ：Step 1: Rewrite the equation \(3x-4y+7=0\) in the slope-intercept form y = mx + b, where m represents the slope and b represents the y-intercept. This yields \(y = \frac{3}{4}x + \frac{7}{4}\). The slope of this line is \(\frac{3}{4}\).

Step 2 ：Step 2: Since parallel lines have equal slopes, the slope of the line we are looking for is also \(\frac{3}{4}\).

Step 3 ：Step 3: Insert the coordinates of the given point (2, -3) into the equation y = mx + b, replacing y with -3, m with \(\frac{3}{4}\), and x with 2. This yields \(-3 = \frac{3}{4}*2 + b\). Solving for b gives \(b = -3 - \frac{3}{4}*2 = -\frac{9}{2}\).

Step 4 ：Step 4: Substitute m = \(\frac{3}{4}\) and b = \(-\frac{9}{2}\) into the equation y = mx + b to get the equation of the line we are looking for, which is \(y = \frac{3}{4}x - \frac{9}{2}\).