JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Straight Lines

The vertex $A$ is the intersection of $3y-x=2$ and $x+y=2$, namely $A(1,1)$; setting $y=0$ in each gives $B(-2,0)$ and $C(2,0)$ on the $x$-axis. Since $BC$ lies along the $x$-axis, the altitude from $A$ is the vertical line $x=1$; the altitude from $B$, perpendicular to $AC$ whose slope is $-1$, is $y=x+2$. These meet at the orthocentre $P(1,3)$. Taking $BC=4$ as base and the ordinate of $P$ as height, the area of triangle $PBC$ is $\frac{1}{2}\cdot 4\cdot 3=6$.