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 $y=-\frac{4}{3}x-\frac{1}{3}$.