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(B) Since $P(1, 1)$ is the circumcentre,it is equidistant from $A, B,$ and $C$. Thus,$PA^2 = PB^2 = PC^2$.$PA^2 = (a-1)^2 + (3-1)^2 = (a-1)^2 + 4$$PB^

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$\frac{4}{7}$

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(B) Since $P(1, 1)$ is the circumcentre,it is equidistant from $A, B,$ and $C$. Thus,$PA^2 = PB^2 = PC^2$.$PA^2 = (a-1)^2 + (3-1)^2 = (a-1)^2 + 4$$PB^2 = (b-1)^2 + (5-1)^2 = (b-1)^2 + 16$$PC^2 = (a-1)^2 + (b-1)^2$Equating $PA^2 = PC^2$: $(a-1)^2 + 4 = (a-1)^2 + (b-1)^2 \implies (b-1)^2 = 4 \implies b-1 = \pm 2$.So $b = 3$ or $b = -1$. Given $ab > 0$,if $b=3$,then $a > 0$. If $b=-1$,then $a Equating $PA^2 = PB^2$: $(a-1)^2 + 4 = (b-1)^2 + 16$.If $b = -1$: $(a-1)^2 + 4 = (-1-1)^2 + 16 = 4 + 16 = 20 \implies (a-1)^2 = 16 \implies a-1 = \pm 4$.$a = 5$ or $a = -3$. Since $a Thus,$A = (-3, 3)$,$B = (-1, 5)$,$C = (-3, -1)$,and $P = (1, 1)$.Line $AP$ passes through $(-3, 3)$ and $(1, 1)$. Slope $m = \frac{1-3}{1-(-3)} = \frac{-2}{4} = -\frac{1}{2}$.Equation of $AP$: $y - 1 = -\frac{1}{2}(x - 1) \implies 2y - 2 = -x + 1 \implies x + 2y = 3$.Line $BC$ passes through $(-1, 5)$ and $(-3, -1)$. Slope $m = \frac{-1-5}{-3-(-1)} = \frac{-6}{-2} = 3$.Equation of $BC$: $y - 5 = 3(x + 1) \implies y = 3x + 8$.Substitute $y$ in $AP$: $x + 2(3x + 8) = 3 \implies 7x + 16 = 3 \implies 7x = -13 \implies x = -\frac{13}{7}$.$y = 3(-\frac{13}{7}) + 8 = \frac{-39 + 56}{7} = \frac{17}{7}$.$k_{1} + k_{2} = -\frac{13}{7} + \frac{17}{7} = \frac{4}{7}$.

Let the circumcentre of a triangle with vertices $A(a, 3)$,$B(b, 5)$,and $C(a, b)$,where $ab > 0$,be $P(1, 1)$. If the line $AP$ intersects the line $BC$ at the point $Q(k_{1}, k_{2})$,then $k_{1} + k_{2}$ is equal to.

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