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},…

Question

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

Aicharya · Accepted Answer

$\mathrm{M}_{\mathrm{AB}}=\frac{4}{-5} \Rightarrow \mathrm{M}_{\mathrm{DP}}=\frac{5}{4}$ 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 $\mathrm{AB}$ $y-1=\frac{5}{4}\left(x-\frac{1}{2}\right)\ldots(4)$ $\begin{aligned} & \text { On solving (3) \& (4) } \\ & (\mathrm{x}-3) 4=5 \mathrm{x}-\frac{5}{2} \\ & \mathrm{x}=\frac{-19}{2}=\mathrm{h} \\ & \mathrm{y}=\frac{-23}{2}=\mathrm{k} \\ & \Rightarrow 2(\mathrm{~h}+\mathrm{k})=-42\end{aligned}$

Two vertices of a triangle $\mathrm{ABC}$ are $\mathrm{A}(3,-1)$ and $\mathrm{B}(-2,3)$, and its orthocentre…

Solution