If the orthocenter of the triangle formed by the lines y = x + 1, y = 4x − 8 and y = mx + c is at (3, −1), then m − c is:

❓ Question

If the orthocenter of the triangle formed by the lines

$$y=x+1,y=4x-8,y=mx+c$$

is at the point $(3,-1)$, then find the value of $m - c$.

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🖼️ Question Image

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✍️ Short Solution

Step 1 — Find intersection of the first two lines (one vertex).
Solve $x+1 = 4x - 8$.
Move terms: $1 + 8 = 4x - x$ → $9 = 3x$ → $x = 3$.
Then $y = x + 1 = 3 + 1 = 4$.
So vertex $A = (3,4)$.

Step 2 — Use the altitude through $A$.
An altitude from vertex $A$ passes through the orthocenter $(3,-1)$. The line through $A(3,4)$ and orthocenter $(3,-1)$ has the same $x$-coordinate, so it is the vertical line $x = 3$. That altitude is vertical, so the side it is perpendicular to must be horizontal. Therefore the third side $y = mx + c$ must be horizontal — i.e. its slope $m = 0$. So the third line is $y = c$.

Step 3 — Use another altitude to find $c$.
Take vertex $B$ as intersection of $y = x + 1$ and $y = c$. Then $x + 1 = c$ ⇒ $x = c - 1$. So $B = (c-1,\,c)$.

The altitude from $B$ must pass through orthocenter $(3,-1)$ and be perpendicular to the opposite side, which is the line $y = 4x - 8$ (slope $4$). The slope of this altitude is the negative reciprocal of $4$, i.e. $-\tfrac{1}{4}$.

So slope between $B(c-1,c)$and orthocenter $(3,-1)$ equals $-\tfrac{1}{4}$:

$$\frac{c-(-1)}{(c-1)-3}=\frac{c+1}{c-4}=-\frac{1}{4}\mathrm{.}$$

Step 4 — Solve for $c$.
Cross-multiply carefully:

$4(c+1) = -1\cdot(c-4)$
Left: $4c + 4$. Right: $-c+4.$

So $4c+4=-c+\mathrm{4\; Move }$$-c$ to left: $4c + c + 4 = 4$ → $5c + 4 = 4$.
Subtract 4 both sides: $5c = 0$. Hence $c = 0$.

We already found $m = 0$, so

$$m-c=0-0=0.$$

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🖼️ Image Solution

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✅ Conclusion & Video Solution

Final answer: $\boxed{m - c = 0}$.

In words: the third line is $y=0$ (a horizontal line) and the algebra of altitudes forces $c=0$, so $m-c=0$.

▶️ Video walkthrough (detailed diagram + verbal explanation):