If the slope of one line is $m$, what is the slope of a line that is perpendicular to it? Apply this to $m = \frac{4}{5}$.

Once you have the slope of line $s$, plug the point $(6,-2)$ into $y = mx + b$ (or use point-slope form) and solve for $b$, which is the $y$-intercept.

In Desmos, enter y = (4/5)x - 4 to represent the given line $4x - 5y = 20$ in slope-intercept form. Note that the coefficient of $x$ is the slope, $\frac{4}{5}$.

In a new line, type y = (-5/4)x + b and let Desmos create a slider for b. This represents all lines with slope $-\frac{5}{4}$, which are perpendicular to the original line.

Type (6,-2) to plot the point the perpendicular line must pass through. Adjust the slider for b until the line y = (-5/4)x + b goes exactly through the point $(6,-2)$. The value of b at that moment is the $y$-intercept; match this value to the correct answer choice.

Put $4x - 5y = 20$ into slope-intercept form.

So the slope of the given line is $m = \frac{4}{5}$.

For two non-vertical lines to be perpendicular, the product of their slopes must be $-1$.

So if the original slope is $\frac{4}{5}$, then the perpendicular slope $m_s$ must satisfy

Solving gives

Thus, line $s$ has slope $-\frac{5}{4}$.

Line $s$ has slope $-\frac{5}{4}$ and passes through $(6,-2)$. Use point-slope form:

which simplifies to

Distribute on the right-hand side:

Now isolate $y$:

The constant term (after simplifying $\frac{15}{2} - 2$) will be the $y$-intercept.

Rewrite $2$ with denominator $2$:

Now subtract:

So the equation of line $s$ is

and the $y$-intercept of line $s$ is $\dfrac{11}{2}$.