JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Straight Lines

Two sides are $3x-2y+6=0$ and $4x+5y-20=0$, meeting at vertex $A$, with orthocentre $H(1,1)$. Each altitude is perpendicular to a side and passes through $H$. Perpendicular to the first side (slope $\frac{3}{2}$) through $H$ gives $2x+3y=5$; perpendicular to the second side (slope $-\frac{4}{5}$) through $H$ gives $5x-4y=1$. Intersecting each altitude with the side it drops onto locates the other two vertices, and the line through them -- the third side -- works out to $26x-122y-1675=0$.