Two vertices of a triangle [ABC] are [A(3,-1)] and [B(-2,3)], and its orthocentre is [P(1,…

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

Accepted Answer

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

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

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)

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

Solutions · 1