JEE Main 2026 — 28 January, Morning Shift — Question 6

$\because \mathrm{m}_{\mathrm{BC}} \cdot \mathrm{m}_{\mathrm{AD}}=-1$

$\Rightarrow\left(-\frac{1}{2 \sqrt{2}}\right)\left(\frac{\beta}{\alpha}\right)=-1$

$$\begin{gathered} \Rightarrow \beta=2 \sqrt{2} \alpha \end{gathered}$$

$\because \mathrm{OD}=\left|\frac{-4}{\sqrt{1+8}}\right|=\frac{4}{3} \Rightarrow \mathrm{AO}=\frac{8}{3}$

So $\mathrm{AD}=\frac{8}{3}+\frac{4}{3}=4$

$\Rightarrow \frac{|\alpha+2 \sqrt{2} \beta-4|}{3}=4 \Rightarrow \alpha=\frac{16}{9}$ or $-\frac{8}{9}$

$\because$ $\mathrm{A}$ $(\alpha, \beta)$ $(0,0)$ lies on same side of given line.

$\therefore(\alpha, \beta)=\left(\frac{16}{9}, \frac{32 \sqrt{2}}{9}\right) ;($ Rejected $)$

So $(\alpha, \beta)=\left(-\frac{8}{9}, \frac{-16 \sqrt{2}}{9}\right)$

$|\alpha+\sqrt{2} \beta|={|\frac{-8-32}{9}|}=4$