JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Straight Lines

For an equilateral triangle the orthocentre coincides with the centroid, so the centroid is the origin. The perpendicular distance from the origin to $BC:x+2\sqrt{2}y=4$ is $\frac{4}{\sqrt{1+8}}=\frac{4}{3}$. The centroid divides the median from $A$ in the ratio $2:1$, so the distance from the centroid to $A$ is $2\times \frac{4}{3}=\frac{8}{3}$, measured along the normal to $BC$ and on the side opposite to $BC$. The unit normal is $\frac{(1,2\sqrt{2})}{3}$, and moving away from the line gives $A=-\frac{8}{3}\cdot \frac{(1,2\sqrt{2})}{3}=\left(-\frac{8}{9},-\frac{16\sqrt{2}}{9}\right)$. Then $\alpha +\sqrt{2}\beta =-\frac{8}{9}-\frac{32}{9}=-\frac{40}{9}$, so $|\alpha +\sqrt{2}\beta |=\frac{40}{9}\approx 4.44$ and the greatest integer not exceeding it is $4$.