QJEEM24Mae92e: JEE Main 2024 Mathematics - Class 11 JEEM24Mae92e Solution: Two vertices of a triangle are and , and its orthocentre is .…

$\begin{aligned}& m_{P A}=\frac{2}{-2}=-1 \\& \therefore \quad m_{B C}=1\end{aligned}$

$\begin{aligned}& B C: y=x+5 \\& m_{B P}=\frac{2}{-3}=\frac{-2}{3} \\& \therefore m_{A C}=\frac{3}{2} \\& A C: y=\frac{3}{2} x-\frac{11}{2} \quad \Rightarrow 2 y=3 x-11 \\& \therefore \quad C:(21,26)\end{aligned}$

Let the circumcentre be $(h, k)$

$\begin{aligned}& (h-1)^2+(k-1)^2=(h+2)^2+(k-3)^2 \quad \text{... (i)}\\& (h-1)^2+(k-1)^2=(h-3)^2+(k+1)^2 \quad \text{... (ii)}\end{aligned}$

Solving (i) and (ii)

$\begin{aligned}& h=\frac{-19}{2}, k=\frac{-23}{2} \\& \alpha+\beta+2(h+k) \\& =21+26-19-23 \\& =2+3=5\end{aligned}$