Find the equation of the line perpendicular to the line $3x-4y=12$ and passes through the point $(2,-1)$.

Step 1 ：First, we convert the given equation into slope-intercept form ($y=mx+b$), where $m$ is the slope and $b$ is the y-intercept. So, $3x-4y=12$ becomes $y=\frac{3}{4}x-3$. Thus, the slope of the given line is $\frac{3}{4}$.

Step 2 ：The slope of a line perpendicular to a line with slope $m$ is $-\frac{1}{m}$. Therefore, the slope of the line we want to find is $-\frac{1}{\frac{3}{4}}=-\frac{4}{3}$.

Step 3 ：Now, we use the point-slope form of the equation of a line, $y-y_1=m(x-x_1)$, where $(x_1,y_1)$ is a point on the line. Substituting the point $(2,-1)$ and the slope $-\frac{4}{3}$ into the equation, we get $y-(-1)=-\frac{4}{3}(x-2)$, which simplifies to $y+1=-\frac{4}{3}x+\frac{8}{3}$.

Step 4 ：Finally, rearranging the equation to slope-intercept form, we get $y=-\frac{4}{3}x+\frac{8}{3}-1=-\frac{4}{3}x+\frac{5}{3}$.