JEE Main 2025 — 7 April, Evening Shift — Question 23

Let the three lines be $L_1: y = x+1$, $L_2: y = 4x-8$, and $L_3: y = mx + c$. The orthocenter is $H = (3, -1)$. Intersection of $L_1$ and $L_2$: $x+1 = 4x-8 \Rightarrow x=3, y=4$.

So vertex $A = (3,4)$. Then vector $\overrightarrow{HA} = (0,5)$.

Since $\overrightarrow{HA}$ is vertical, the opposite side $BC$ must be horizontal.

Thus the y-coordinates of B and C are equal: $y_B = y_C$. From the intersection of $L_1$ and $L_3$: $x_B = \frac{c-1}{1-m}, y_B = \frac{c-m}{1-m}$. From the intersection of $L_2$ and $L_3$: $x_C = \frac{c+8}{4-m}, y_C = \frac{4c+8m}{4-m}$. Setting $y_B = y_C$ gives $\frac{c-m}{1-m} = \frac{4c+8m}{4-m}$.

Cross-multiplying and simplifying yields $m(c-4+3m)=0$. Now use the altitude from $B$: $\overrightarrow{HB} \perp \overrightarrow{AC}$. First, if $m eq 0$, then $c=4-3m$ and $y_B = y_C = 4$ (from solving further).

But then all three vertices lie on $y=4$, making the triangle degenerate.

So $m=0$ is the only possibility. With $m=0$, $L_3$ is $y=c$. Then $B = (c-1, c)$, $C = \left(\frac{c+8}{4}, c\right)$. Compute $\overrightarrow{HB} = (c-4, c+1)$ and $\overrightarrow{AC} = \left(\frac{c-4}{4}, c-4\right)$. Dot product: $(c-4)\cdot\frac{c-4}{4} + (c+1)(c-4) = \frac{(c-4)^2}{4} + (c+1)(c-4) = 0$. Factor $(c-4)$: $(c-4)\left(\frac{c-4}{4} + (c+1)\right) = 0$. Simplify: $(c-4)\left(\frac{5c}{4}\right) = 0 \Rightarrow c=0 \text{ or } c=4$. $c=4$ makes all lines concurrent at $(3,4)$ (degenerate). Hence $c=0$. Thus $m=0$, $c=0$, and $m-c = 0$.