Coordinate Geometry – Straight Lines – JEE Main 28 Jan 2026 Shift 1

Solution:

Let $O(0,0)$ be the orthocenter (which is also the centroid in an equilateral triangle).
The equation of the line BC is $x + 2\sqrt{2}y – 4 = 0$.
Let AD be the altitude from A to BC, intersecting BC at D.
The length of the perpendicular from the origin O to the line BC is:
$$OD = \frac{|0 + 2\sqrt{2}(0) – 4|}{\sqrt{1^2 + (2\sqrt{2})^2}} = \frac{4}{\sqrt{1 + 8}} = \frac{4}{3}$$

Since O is the centroid, it divides the altitude AD in the ratio $2:1$. Thus, $AO = 2 \cdot OD$.
$$AO = 2 \times \frac{4}{3} = \frac{8}{3}$$

The line AD passes through the origin and is perpendicular to BC.
Slope of BC is $m_{BC} = -\frac{1}{2\sqrt{2}}$.
Slope of AD is $m_{AD} = -\frac{1}{m_{BC}} = 2\sqrt{2}$.
Equation of line OA (which is same as AD) is $y = 2\sqrt{2}x$, so $\beta = 2\sqrt{2}\alpha$.

Distance $AO = \frac{8}{3}$:
$$\sqrt{\alpha^2 + \beta^2} = \frac{8}{3}$$
Substituting $\beta = 2\sqrt{2}\alpha$:
$$\sqrt{\alpha^2 + (2\sqrt{2}\alpha)^2} = \frac{8}{3}$$
$$\sqrt{\alpha^2 + 8\alpha^2} = \frac{8}{3}$$
$$\sqrt{9\alpha^2} = \frac{8}{3} \Rightarrow 3|\alpha| = \frac{8}{3} \Rightarrow |\alpha| = \frac{8}{9}$$

From the line equation $x + 2\sqrt{2}y = 4$, the intercept is in the first quadrant. Since A and D are on opposite sides of the origin, A lies in the third quadrant.
So, $\alpha = -\frac{8}{9}$ and $\beta = 2\sqrt{2}(-\frac{8}{9}) = -\frac{16\sqrt{2}}{9}$.

We need to find the value of $|\alpha + \sqrt{2}\beta|$:
$$|\alpha + \sqrt{2}\beta| = \left| -\frac{8}{9} + \sqrt{2}\left(-\frac{16\sqrt{2}}{9}\right) \right|$$
$$= \left| -\frac{8}{9} – \frac{32}{9} \right| = \left| -\frac{40}{9} \right| = \frac{40}{9} = 4.44…$$

The greatest integer less than or equal to this value is:
$$[\frac{40}{9}] = 4$$

Ans. (4)