JEE Main 2026 (Online) 5th April Morning ShiftQuestion 1
In an equilateral triangle $PQR$, let the vertex $P$ be at $(3,5)$ and the side $QR$ be along the line $x+y=4$. If the orthocentre of the triangle $PQR$ is $(\alpha,\beta)$, then $9(\alpha+\beta)$ is equal to :
Points and Straight Lines
Solution
Key fact. In an equilateral triangle the orthocentre, centroid, incentre and circumcentre coincide. So the orthocentre is the centroid $G$, which lies on the altitude $PM$ and divides it as $PG:GM=2:1$.Step 1 - Foot of the altitude. $QR$ lies on $x+y-4=0$. The foot $M$ of the perpendicular from $P(3,5)$ is $$M=P-\frac{(3)+(5)-4}{1^2+1^2}\,(1,1)=(3,5)-2(1,1)=(1,3).$$ Step 2 - The centroid. $G$ is two-thirds of the way from $P$ to $M$: $$G=P+\tfrac23(M-P)=(3,5)+\tfrac23(-2,-2)=\left(\tfrac53,\tfrac{11}{3}\right).$$ Step 3 - Value asked. $$9(\alpha+\beta)=9\left(\tfrac53+\tfrac{11}{3}\right)=9\cdot\tfrac{16}{3}=48 .$$ Answer: (D)