Two vertices of a triangle ABC are A(3, -1) a | vedclass

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(C) The slope of $AB$ is $M_{AB} = \frac{3 - (-1)}{-2 - 3} = \frac{4}{-5}$.Since $CP \perp AB$,the slope of $CP$ is $M_{CP} = \frac{5}{4}$.The equatio

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$5$

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(C) The slope of $AB$ is $M_{AB} = \frac{3 - (-1)}{-2 - 3} = \frac{4}{-5}$.Since $CP \perp AB$,the slope of $CP$ is $M_{CP} = \frac{5}{4}$.The equation of line $CP$ passing through $P(1, 1)$ is $y - 1 = \frac{5}{4}(x - 1) \Rightarrow 5x - 4y - 1 = 0$ ... $(1)$.The slope of $AP$ is $M_{AP} = \frac{1 - (-1)}{1 - 3} = \frac{2}{-2} = -1$.Since $BC \perp AP$,the slope of $BC$ is $M_{BC} = 1$.The equation of line $BC$ passing through $B(-2, 3)$ is $y - 3 = 1(x + 2) \Rightarrow x - y + 5 = 0$ ... $(2)$.Solving $(1)$ and $(2)$: From $(2)$,$y = x + 5$. Substituting into $(1)$: $5x - 4(x + 5) - 1 = 0$ $\Rightarrow 5x - 4x - 20 - 1 = 0$ $\Rightarrow x = 21$. Thus $\alpha = 21$.Then $\beta = 21 + 5 = 26$. So,$\alpha + \beta = 21 + 26 = 47$.For the circumcentre $(h, k)$ of $	riangle PAB$,note that $P$ is the orthocentre of $	riangle ABC$. $A$ known property is that the circumcircle of $	riangle PAB$ has the same radius as the circumcircle of $	riangle ABC$,and its centre is the reflection of the circumcentre of $	riangle ABC$ across $AB$. Alternatively,the circumcentre of $	riangle PAB$ is the point $O'$ such that $O'A = O'B = O'P$. The perpendicular bisector of $AB$ is $y - 1 = \frac{5}{4}(x - 0.5)$. The perpendicular bisector of $AP$ is $y - 0 = 1(x - 2)$. Solving these gives $h = -19/2$ and $k = -23/2$.Thus,$2(h + k) = 2(-19/2 - 23/2) = -42$.The final value is $(\alpha + \beta) + 2(h + k) = 47 - 42 = 5$.

Two vertices of a triangle $ABC$ are $A(3, -1)$ and $B(-2, 3)$,and its orthocentre is $P(1, 1)$. If the coordinates of the point $C$ are $(\alpha, \beta)$ and the centre of the circle circumscribing the triangle $PAB$ is $(h, k)$,then the value of $(\alpha + \beta) + 2(h + k)$ equals :

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