Find the equation of the line parallel to the line $3x-4y=12$ and passing through the point $(2,-3)$.

Step 1 ：Step 1: Convert the given line to slope-intercept form ($y=mx+b$), where $m$ is the slope. So, we have: $$3x-4y=12\Rightarrow y=\frac{3}{4}x-3$$ The slope of this line is $\frac{3}{4}$.

Step 2 ：Step 2: A line parallel to another line will have the same slope. So, the slope of the line we are trying to find is also $\frac{3}{4}$.

Step 3 ：Step 3: We know the slope ($m$) and a point on the line ($(2,-3)$). We can plug these values into the point-slope form of a line equation: $$y-y_1=m(x-x_1)$$ where $(x_1,y_1)$ are the coordinates of the known point. Substituting for $m$, $x_1$, and $y_1$, we get: $$y-(-3)=\frac{3}{4}(x-2)$$

Step 4 ：Step 4: Simplifying the equation, we get: $$y+3=\frac{3}{4}x-\frac{3}{2}$$ $$y=\frac{3}{4}x-\frac{3}{2}-3$$ $$y=\frac{3}{4}x-\frac{9}{2}$$