Key Concepts and Formulas

• Orthocenter: The point of intersection of the altitudes of a triangle.
• Altitude: A line segment from a vertex of a triangle perpendicular to the opposite side.
• Perpendicular Lines: If two lines have slopes $m_1$ and $m_2$, and they are perpendicular, then $m_1 m_2 = -1$.
• Equation of a Line: A line with slope $m$ passing through the point $(x_1, y_1)$ has the equation $y - y_1 = m(x - x_1)$.

Step-by-Step Solution

Step 1: Identify the given lines and the orthocenter.

We are given the equations of three lines: $L_1: y = x + 1$ $L_2: y = 4x - 8$ $L_3: y = mx + c$ The orthocenter $H$ is given as $(3, -1)$.

Step 2: Find the slopes of the given lines.

The slopes of the lines are: $m_1 = 1 \text{ (for } L_1)$ $m_2 = 4 \text{ (for } L_2)$ $m_3 = m \text{ (for } L_3)$

Step 3: Determine the equations of the altitudes from vertices to the opposite sides.

Let $A$ be the vertex formed by the intersection of $L_1$ and $L_2$. Let $B$ be the vertex formed by the intersection of $L_1$ and $L_3$. Let $C$ be the vertex formed by the intersection of $L_2$ and $L_3$.

The altitude from $A$ is perpendicular to $L_3$, so its slope is $-\frac{1}{m}$. The altitude from $B$ is perpendicular to $L_2$, so its slope is $-\frac{1}{4}$. The altitude from $C$ is perpendicular to $L_1$, so its slope is $-1$.

Since the orthocenter is the intersection of the altitudes, we know that the altitudes pass through $(3, -1)$.

Step 4: Find the equation of the altitude from C to $L_1$.

The altitude from $C$ has a slope of $-1$ and passes through $(3, -1)$. Therefore, its equation is: $y - (-1) = -1(x - 3)$ $y + 1 = -x + 3$ $y = -x + 2$

Since the altitude from $C$ is perpendicular to $L_1$, the intersection point $C$ must lie on $L_2$ and $L_3$. It also lies on the altitude. The intersection of $L_2$ and the altitude from $C$: $4x - 8 = -x + 2$ $5x = 10$ $x = 2$ $y = -2 + 2 = 0$ So, the coordinates of $C$ are $(2, 0)$.

Step 5: Find the equation of the altitude from B to $L_2$.

The altitude from $B$ has a slope of $-\frac{1}{4}$ and passes through $(3, -1)$. Therefore, its equation is: $y - (-1) = -\frac{1}{4}(x - 3)$ $y + 1 = -\frac{1}{4}x + \frac{3}{4}$ $y = -\frac{1}{4}x - \frac{1}{4}$

Since the altitude from $B$ is perpendicular to $L_2$, the intersection point $B$ must lie on $L_1$ and $L_3$. It also lies on the altitude. The intersection of $L_1$ and the altitude from $B$: $x + 1 = -\frac{1}{4}x - \frac{1}{4}$ $\frac{5}{4}x = -\frac{5}{4}$ $x = -1$ $y = -1 + 1 = 0$ So, the coordinates of $B$ are $(-1, 0)$.

Step 6: Find the equation of $L_3$.

Since $L_3$ passes through points $B(-1, 0)$ and $C(2, 0)$, its equation is simply $y=0$. Comparing this to $y = mx + c$, we have $m=0$ and $c=0$. Therefore, $m-c = 0-0 = 0$.

Step 7: Verify that the orthocenter lies on the altitude from A.

The vertex A is found by intersecting $L_1$ and $L_2$: $x+1 = 4x-8$ $3x = 9$ $x=3$ $y=3+1=4$ So $A = (3,4)$.

The altitude from A has slope $-\frac{1}{m}$, which is undefined since $m=0$. This means the altitude from A is a vertical line $x=3$. Since the orthocenter is $(3,-1)$, it indeed lies on this altitude.

Common Mistakes & Tips

• Be careful when calculating the slopes of perpendicular lines. The product of their slopes must be -1.
• Remember that the orthocenter lies on the altitudes, not necessarily on the sides of the triangle.
• When dealing with vertical or horizontal lines, remember that their slopes are undefined or zero, respectively.

Summary

We found the equations of the altitudes from two vertices and used them to determine the coordinates of those vertices. The coordinates of vertices B and C directly gave us the equation of the third side $L_3$ as $y=0$. Thus, $m=0$ and $c=0$ and $m-c = 0$.

Final Answer The final answer is \boxed{0}, which corresponds to option (A).