JEE Main 2024 — Mathematics Coordinate Geometry

$\mathrm{M}_{\mathrm{AB}}=\frac{4}{-5} \Rightarrow \mathrm{M}_{\mathrm{DP}}=\frac{5}{4}$MAB​=−54​⇒MDP​=45​ Equation of $\mathrm{PC}$PC is $\mathrm{y}-1=\frac{5}{4}(\mathrm{x}-1) \ldots(1)$y−1=45​(x−1)…(1) $\mathrm{M}_{\mathrm{AP}}=\frac{2}{-2}=-1 \Rightarrow \mathrm{M}_{\mathrm{BC}}=+1$MAP​=−22​=−1⇒MBC​=+1 Equation of $\mathrm{BC}$BC is $\mathrm{y}-3=(\mathrm{x}+2)\ldots(2)$y−3=(x+2)…(2) On solving (1) and (2) $\begin{aligned}
& x+4=\frac{5}{4}(x-1) \Rightarrow 4 x+16=5 x-5 \Rightarrow \alpha=21 \
& \Rightarrow \beta=y=x+5=26 \
& \alpha+\beta=47
\end{aligned}$Equation of$Equationof\perp$bisector of AP$bisectorofAPy-0=(x-2)\ldots(3)$Equation of$Equationof\perp$bisector of$bisectorof\mathrm{AB}$$y-1=\frac{5}{4}\left(x-\frac{1}{2}\right)\ldots(4)$$\begin{aligned} & \text { On solving (3) & (4) } \ & (\mathrm{x}-3) 4=5 \mathrm{x}-\frac{5}{2} \ & \mathrm{x}=\frac{-19}{2}=\mathrm{h} \ & \mathrm{y}=\frac{-23}{2}=\mathrm{k} \ & \Rightarrow 2(\mathrm{~h}+\mathrm{k})=-42\end{aligned}$