JEE Main 2026 — 22 January, Evening Shift — Question 7

Solution of statement-1 $\mathrm{m}_{\mathrm{AO}} \cdot \mathrm{m}_{\mathrm{BC}}=-1$

$$\begin{gathered} \Rightarrow 5 \mathrm{~h}-\mathrm{k}+13=0 \end{gathered}$$

$\mathrm{m}_{\mathrm{AB}} \cdot \mathrm{m}_{\mathrm{OC}}=-1$

$\begin{gathered} \Rightarrow 4 \mathrm{k}=7 \mathrm{~h} \end{gathered}$ ⇒ third vertex is $(-4,-7)$

∴ Statement 1 is correct.

Solution of statement-2

$2 \mathrm{a}, \mathrm{b}, \mathrm{c} \rightarrow$ A.P. $\mathrm{b}=\frac{2 \mathrm{a}+\mathrm{c}}{2}$

$\Rightarrow 2 \mathrm{a}-2 \mathrm{~b}+\mathrm{c}=0$

∵ lines $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ are concurrent then $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{-2}=\frac{1}{1}$

$\mathrm{x}=2$ and $\mathrm{y}=-2$

∴ Point of concurrency is $(2,-2)$

∴ Statement 2 is correct.