JEE Main 2026 — Straight Lines Question with Solution

In an equilateral triangle, the orthocentre coincides with the centroid.

The centroid divides the altitude from the vertex to the opposite side in the ratio $2:1$.

Let $M$ be the foot of the perpendicular from $P(3,5)$ to the line $QR$ given by $x+y-4=0$. The coordinates of $M$ can be found using the formula:

$\frac{x-3}{1}=\frac{y-5}{1}=-\frac{3+5-4}{1^2+1^2}$

$\frac{x-3}{1}=\frac{y-5}{1}=-2$

This gives $x=1$ and $y=3$. Thus, $M$ is $(1,3)$.

The orthocentre $H(\alpha ,\beta )$ divides the segment $PM$ internally in the ratio $2:1$. Using the section formula:

$\alpha =\frac{2(1)+1(3)}{2+1}=\frac{5}{3}$

$\beta =\frac{2(3)+1(5)}{2+1}=\frac{11}{3}$

Adding $\alpha$ and $\beta$:

$\alpha +\beta =\frac{5}{3}+\frac{11}{3}=\frac{16}{3}$

Therefore, $9(\alpha +\beta )=9\left(\frac{16}{3}\right)=48$

Answer: $48$