JEE Main 2025 — 24 January, Morning Shift — Question 19

$\text{Let\; the\; line\; be\; }2x-3y+\lambda=0.$

$\text{For\; }(x_0,y_0)=(1,2)\;\text{ the\; reflection\; formula\; gives\; }d=\frac{ax_0+by_0+c}{a^2+b^2}.$

$a=2,\ b=-3,\ c=\lambda\Rightarrow d=\frac{2(1)-3(2)+\lambda}{4+9}=\frac{\lambda-4}{13}.$

$\text{The\; reflected\; point\; is\; }(x',y')=(x_0-2ad,\;y_0-2bd).$

$\Rightarrow x'=1-4d,\quad y'=2+6d.$

$\text{Given\; }(x',y')=\big(\tfrac{57}{13},-\tfrac{40}{13}\big).$

$1-4d=\tfrac{57}{13}\Rightarrow 4d=1-\tfrac{57}{13}=-\tfrac{44}{13}\Rightarrow d=-\tfrac{11}{13}.$

$\text{Thus\; } \frac{\lambda-4}{13}=-\frac{11}{13}\Rightarrow \lambda=-7.$

$\text{Now\; concurrency:\; solve\; } \begin{cases}8x-11y=33\\[4pt]2x-3y=7\end{cases}$

$\text{Subtract\; }(8x-12y=28)\; \text{ from }(8x-11y=33)\Rightarrow y=5,\; x=11.$

$\text{Plug\; into\; }3x-4y-\alpha=0:\;33-20-\alpha=0\Rightarrow \alpha=13.$

$|\alpha\lambda|=|13\cdot(-7)|=91.$

$\boxed{91}$