1. Let the three lines forming the triangle be: $L_1: y=x+1,\quad L_2: y=4x-8,\quad L_3: y=mx+c$

   The orthocenter of this triangle is given as $H=(3,-1).$

2. In a triangle, the altitude from a vertex passes through the orthocenter and is perpendicular to the opposite side.

   So we first find two vertices formed by $L_1$ and $L_2$ with each other.

3. Vertex $A=L_1\cap L_2$: $x+1=4x-8$ $3x=9\Rightarrow x=3$ $y=4$ Hence, $A=(3,4).$

4. Since $A$ is the intersection of $L_1$ and $L_2$, the opposite side is $L_3$.

   The altitude from $A$ must pass through $H=(3,-1)$. Since both points have the same $x$-coordinate, the line $AH$ is $x=3,$ which is a vertical line.

   Therefore, the opposite side $L_3$ must be horizontal.

   Hence its slope is $m=0.$

   So, $L_3: y=c.$

5. Now find the other two vertices:

   • $B=L_2\cap L_3$ lies on $y=4x-8$ and $y=c$.
   • $C=L_1\cap L_3$ lies on $y=x+1$ and $y=c$.

   Since $L_3$ is horizontal, the altitude from $B$ is perpendicular to $L_1$.

   Slope of $L_1$ is $1$, so slope of altitude from $B$ is $-1$. This altitude passes through $H=(3,-1)$, hence its equation is $y+1=-1(x-3)$ $y=-x+2.$

6. Point $B$ is the intersection of this altitude with $L_2: y=4x-8$. So, $4x-8=-x+2$ $5x=10\Rightarrow x=2$ $y=0.$

   Thus, $B=(2,0).$

   Since $B$ also lies on $L_3: y=c$, we get $c=0.$

7. Therefore, $m-c=0-0=0.$

8. Checking options:

   • A: $0$ ✅
   • B: $2$
   • C: $-2$
   • D: $4$

Therefore the correct answer is $\boxed{0}.$