JEE MAIN - Mathematics (2024 - 9th April Evening Shift - No. 5)
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
81
15
51
5
Explanation
$$\begin{aligned} & m_{P A}=\frac{2}{-2}=-1 \\ & \therefore \quad m_{B C}=1 \end{aligned}$$
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$$\begin{aligned} & B C: y=x+5 \\ & m_{B P}=\frac{2}{-3}=\frac{-2}{3} \\ & \therefore m_{A C}=\frac{3}{2} \\ & A C: y=\frac{3}{2} x-\frac{11}{2} \quad \Rightarrow 2 y=3 x-11 \\ & \therefore \quad C:(21,26) \end{aligned}$$
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)
$$\begin{aligned} & h=\frac{-19}{2}, k=\frac{-23}{2} \\ & \alpha+\beta+2(h+k) \\ & =21+26-19-23 \\ & =2+3=5 \end{aligned}$$
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