JEE Main 2025 Apr 7 Shift 2, Mathematics Q1: Introduction to Straight Lines

The lines $y=x+1$ and $y=4x-8$ meet where $x+1=4x-8$, giving $x=3,y=4$, so vertex $P=(3,4)$.

Since the orthocentre $H=(3,-1)$ also has $x$-coordinate 3, the altitude from $P$ is the vertical line $x=3$. An altitude is perpendicular to the side it drops onto, so the side opposite $P$ (the line $y=mx+c$) must be horizontal, i.e. $m=0$. This side then has equation $y=c$.

With $m=0$, the other two vertices are $Q$ (intersection of $y=x+1$ with $y=c$) $=(c-1,c)$ and $R$ (intersection of $y=4x-8$ with $y=c$) $=\left(\frac{c+8}{4},c\right)$.

The altitude from $Q$ must be perpendicular to side $PR$, which lies along $y=4x-8$ (slope 4), so this altitude has slope $-\frac{1}{4}$ and passes through $Q$ and $H(3,-1)$:

$\frac{-1-c}{3-(c-1)}=-\frac{1}{4}\Rightarrow 4(-1-c)=-(4-c)\Rightarrow -4-4c=-4+c\Rightarrow c=0$.

$\therefore m-c=0-0=0$.