The slope of a line perpendicular to $-\frac{3}{4}$ is the negative reciprocal. Use that slope and the midpoint of $\overline{PQ}$ to write an equation for line $p$.

Solve the system formed by line $m$ and line $p$ (not line $m$ and line $n$), then plug the solution into $12y-6x$.

Enter 5x-2y=4.

Plot the points $P(0,8)$ and $Q(8,2)$, then enter the line through them (for example, y=(-3/4)x+8).

Compute the midpoint $M=(4,5)$. Since line $n$ has slope $-3/4$, a perpendicular line has slope $4/3$. Enter y-5=(4/3)(x-4) to graph line $p$.

Click the intersection of lines $m$ and $p$ to get $(x,y)$. Then evaluate 12y-6x by substituting those coordinate values (use parentheses).

From the graph, $P(0,8)$ and $Q(8,2)$ lie on line $n$.

The slope of $n$ is

The midpoint of $\overline{PQ}$ is

A line perpendicular to slope $-\frac{3}{4}$ has slope $\frac{4}{3}$ (the negative reciprocal).

Using point-slope form through $(4,5)$:

In slope-intercept form,

Rewrite line $m$ in slope-intercept form:

Set the two expressions for $y$ equal:

Multiply by $6$:

Then

Substitute $x=\frac{10}{7}$ and $y=\frac{11}{7}$:

Therefore, the value of $12y-6x$ is $\frac{72}{7}$.