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Two vertices of a triangle $\mathrm{ABC}$ are $\mathrm{A}(3,-1)$ and $\mathrm{B}(-2,3)$, and its orthocentre is $\mathrm{P}(1,1)$. If the coordinates of the point $\mathrm{C}$ are $(\alpha, \beta)$ and the centre of the of the circle circumscribing the triangle $\mathrm{PAB}$ is $(\mathrm{h}, \mathrm{k})$, then the value of $(\alpha+\beta)+2(\mathrm{~h}+\mathrm{k})$ equals
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The correct answer is:
5
$\mathrm{M}_{\mathrm{AB}}=\frac{4}{-5} \Rightarrow \mathrm{M}_{\mathrm{DP}}=\frac{5}{4}$
Equation of $\mathrm{PC}$ is $\mathrm{y}-1=\frac{5}{4}(\mathrm{x}-1) \ldots(1)$
$\mathrm{M}_{\mathrm{AP}}=\frac{2}{-2}=-1 \Rightarrow \mathrm{M}_{\mathrm{BC}}=+1$
Equation of $\mathrm{BC}$ is $\mathrm{y}-3=(\mathrm{x}+2)\ldots(2)$
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 $\perp$ bisector of AP
$y-0=(x-2)\ldots(3)$
Equation of $\perp$ bisector of $\mathrm{AB}$
$y-1=\frac{5}{4}\left(x-\frac{1}{2}\right)\ldots(4)$
$\begin{aligned} & \text { On solving (3) \& (4) } \\ & (\mathrm{x}-3) 4=5 \mathrm{x}-\frac{5}{2} \\ & \mathrm{x}=\frac{-19}{2}=\mathrm{h} \\ & \mathrm{y}=\frac{-23}{2}=\mathrm{k} \\ & \Rightarrow 2(\mathrm{~h}+\mathrm{k})=-42\end{aligned}$
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