Animated Solution for Mathematics - Straight Lines: If the orthocentre 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 :

Welcome, fellow traveler on the path to JEE excellence. Today, we are not just solving a coordinate geometry problem; we are peeling back the layers of a triangle to reveal the hidden elegance of its altitudes.

Imagine you are standing on a coordinate plane. You have two lines, $L_1:y=x+1$ and $L_2:y=4x-8$, intersecting at a point $P$. You are also given a mysterious point $H(3,-1)$, the orthocenter. Our goal is to find the third line $L_3:y=mx+c$ that completes this triangle.

The First Vertex and the Vertical Altitude

Let us begin by finding the vertex $P$. By equating the two known lines, $x+1=4x-8$, we find $3x=9$, which gives us $x=3$.

Substituting this back, we get $y=4$. Thus, our first vertex is $P(3,4)$.

Now, look closely at the orthocenter $H(3,-1)$. Both $P$ and $H$ share the same x-coordinate of $3$. This is a gift from the geometry gods! It means the altitude passing through $P$ and $H$ is a vertical line, $x=3$.

The Horizontal Constraint

If the altitude from $P$ is vertical, then the side $L_3$ to which it is perpendicular must be horizontal. A horizontal line has a slope of $m=0$.

This simplifies our third line significantly: $L_3$ is now just $y=c$. We have already slashed the complexity of our problem by half. We know $m=0$, and now we only need to find $c$.

The Intersection and the Slope Condition

To find $c$, we need another vertex. Let's call the intersection of $L_1$ and $L_3$ vertex $Q$. Since $L_1$ is $y=x+1$ and $L_3$ is $y=c$, we find $x=c-1$. So, $Q$ is $(c-1,c)$.

Now, we invoke the definition of the orthocenter again. The altitude from $Q$ must pass through $H(3,-1)$ and be perpendicular to the side $L_2$.

The slope of $L_2$ is $4$. Therefore, the slope of the altitude $QH$ must be $-\frac{1}{4}$. We calculate the slope of $QH$ using the coordinates $Q(c-1,c)$ and $H(3,-1)$:

The Final Convergence

Now, we set this slope equal to $-\frac{1}{4}$:

Cross-multiplying gives us $4(c+1)=-1(c-4)$, which simplifies to $4c+4=-c+4$. The constants cancel out beautifully, leaving us with $5c=0$, or $c=0$.

We have arrived! We found $m=0$ and $c=0$. The question asks for $m-c$, which is $0-0=0$.

It is a moment of pure satisfaction when the variables vanish, leaving behind the elegant truth of the answer. The final result is 0. Remember, in JEE Advanced, the math is not just about calculation; it is about recognizing the geometric constraints that simplify the path.