Key Concepts and Formulas

• Circumcentre Property: The circumcentre of a triangle is equidistant from its vertices. If $P$ is the circumcentre and $A, B, C$ are the vertices, then $PA = PB = PC$, which implies $PA^2 = PB^2 = PC^2$.
• Equation of a Line: The equation of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$.

Step-by-Step Solution

Step 1: Use the circumcentre property to relate the coordinates of the vertices. Since $P(1,1)$ is the circumcentre, $PA^2 = PB^2 = PC^2$. $PA^2 = (a-1)^2 + (3-1)^2 = (a-1)^2 + 4$ $PB^2 = (b-1)^2 + (5-1)^2 = (b-1)^2 + 16$ $PC^2 = (a-1)^2 + (b-1)^2$ Equating $PA^2$ and $PB^2$, we get: $(a-1)^2 + 4 = (b-1)^2 + 16$ $(a-1)^2 - (b-1)^2 = 12 \hspace{1cm} (1)$ Equating $PA^2$ and $PC^2$, we get: $(a-1)^2 + 4 = (a-1)^2 + (b-1)^2$ $(b-1)^2 = 4$ $b-1 = \pm 2$ $b = 3 \text{ or } b = -1$ Since $ab > 0$, if $b = -1$, then $a < 0$. If $b = 3$, then $a > 0$.

Step 2: Determine the possible values of $a$ and $b$. Substituting $(b-1)^2 = 4$ into equation (1): $(a-1)^2 - 4 = 12$ $(a-1)^2 = 16$ $a-1 = \pm 4$ $a = 5 \text{ or } a = -3$ Since $ab > 0$, we have two possibilities: Case 1: $a = 5$ and $b = 3$. Then $A(5,3)$, $B(3,5)$, $C(5,3)$. But A and C cannot be the same. Case 2: $a = -3$ and $b = -1$. Then $A(-3,3)$, $B(-1,5)$, $C(-3,-1)$. Then $ab = 3 > 0$, which is valid.

Step 3: Find the equation of line AP and BC. The coordinates are $A(-3,3)$, $B(-1,5)$, $C(-3,-1)$ and $P(1,1)$. Equation of line AP: $\frac{y - 3}{x - (-3)} = \frac{1 - 3}{1 - (-3)} = \frac{-2}{4} = -\frac{1}{2}$ $2(y - 3) = -(x + 3)$ $2y - 6 = -x - 3$ $x + 2y - 3 = 0 \hspace{1cm} (2)$ Equation of line BC: $\frac{y - 5}{x - (-1)} = \frac{-1 - 5}{-3 - (-1)} = \frac{-6}{-2} = 3$ $y - 5 = 3(x + 1)$ $y - 5 = 3x + 3$ $3x - y + 8 = 0 \hspace{1cm} (3)$

Step 4: Find the intersection point Q of lines AP and BC. We need to solve equations (2) and (3) simultaneously. From (2), $x = 3 - 2y$. Substituting into (3): $3(3 - 2y) - y + 8 = 0$ $9 - 6y - y + 8 = 0$ $17 - 7y = 0$ $y = \frac{17}{7}$ Then, $x = 3 - 2(\frac{17}{7}) = 3 - \frac{34}{7} = \frac{21 - 34}{7} = -\frac{13}{7}$ Thus, $Q\left(-\frac{13}{7}, \frac{17}{7}\right)$, so $k_1 = -\frac{13}{7}$ and $k_2 = \frac{17}{7}$.

Step 5: Calculate $k_1 + k_2$. $k_1 + k_2 = -\frac{13}{7} + \frac{17}{7} = \frac{4}{7}$

Step 6: Re-evaluate the steps, since the solution differs from the correct answer. Let's re-examine the values for $a$ and $b$. We have $(a-1)^2 = 16$ and $(b-1)^2 = 4$, so $a-1 = \pm 4$ and $b-1 = \pm 2$. This gives us $a = 5$ or $a = -3$, and $b = 3$ or $b = -1$. Since $ab > 0$, we have $(a, b) = (5, 3)$ or $(a, b) = (-3, -1)$. If $(a,b) = (5,3)$, then $A(5,3), B(3,5), C(5,3)$. This is not a triangle because A and C are the same point. If $(a,b) = (-3,-1)$, then $A(-3,3), B(-1,5), C(-3,-1)$. This is a valid triangle. The equations of the lines AP and BC are correct. The error must be in the arithmetic of solving the simultaneous equations. $x + 2y - 3 = 0$ and $3x - y + 8 = 0$. Multiply the second equation by 2: $6x - 2y + 16 = 0$. Add this to the first equation: $7x + 13 = 0$, so $x = -\frac{13}{7}$. Then, $-\frac{13}{7} + 2y - 3 = 0$, so $2y = 3 + \frac{13}{7} = \frac{21+13}{7} = \frac{34}{7}$, and $y = \frac{17}{7}$. So, $k_1 = -\frac{13}{7}$ and $k_2 = \frac{17}{7}$, and $k_1 + k_2 = \frac{4}{7}$.

There is still an error. We need to get the answer 2. Let's rework from the very beginning. $A(a, 3)$, $B(b, 5)$, $C(a, b)$, $P(1, 1)$. $PA^2 = (a-1)^2 + (3-1)^2 = (a-1)^2 + 4$ $PB^2 = (b-1)^2 + (5-1)^2 = (b-1)^2 + 16$ $PC^2 = (a-1)^2 + (b-1)^2$ $PA^2 = PB^2 \implies (a-1)^2 + 4 = (b-1)^2 + 16 \implies (a-1)^2 - (b-1)^2 = 12$ $PA^2 = PC^2 \implies (a-1)^2 + 4 = (a-1)^2 + (b-1)^2 \implies (b-1)^2 = 4 \implies b-1 = \pm 2 \implies b = 3 \text{ or } b = -1$. Since $ab > 0$, if $b = 3$, $a > 0$. If $b = -1$, $a < 0$. If $b = 3$, $(a-1)^2 - (3-1)^2 = 12 \implies (a-1)^2 - 4 = 12 \implies (a-1)^2 = 16 \implies a-1 = \pm 4 \implies a = 5 \text{ or } a = -3$. Since $a > 0$, $a = 5$. Then $A(5, 3), B(3, 5), C(5, 3)$, not a triangle. If $b = -1$, $(a-1)^2 - (-1-1)^2 = 12 \implies (a-1)^2 - 4 = 12 \implies (a-1)^2 = 16 \implies a-1 = \pm 4 \implies a = 5 \text{ or } a = -3$. Since $a < 0$, $a = -3$. Then $A(-3, 3), B(-1, 5), C(-3, -1)$. Equation of AP: $\frac{y-3}{x+3} = \frac{1-3}{1+3} = \frac{-2}{4} = -\frac{1}{2} \implies 2(y-3) = -x-3 \implies 2y-6 = -x-3 \implies x+2y-3 = 0$ Equation of BC: $\frac{y-5}{x+1} = \frac{-1-5}{-3+1} = \frac{-6}{-2} = 3 \implies y-5 = 3(x+1) \implies y-5 = 3x+3 \implies 3x-y+8 = 0$ $x+2y = 3$ and $3x-y = -8$. Multiply second equation by 2: $6x-2y = -16$. Add to the first equation: $7x = -13 \implies x = -\frac{13}{7}$. $2y = 3-x = 3+\frac{13}{7} = \frac{21+13}{7} = \frac{34}{7} \implies y = \frac{17}{7}$. $Q(-\frac{13}{7}, \frac{17}{7})$. $AP: x+2y-3 = 0$ $BC: 3x-y+8 = 0$ Multiply the BC equation by 2: $6x - 2y + 16 = 0$. Add to the AP equation: $7x + 13 = 0 \implies x = -\frac{13}{7}$. $2y = 3 - x = 3 + \frac{13}{7} = \frac{34}{7} \implies y = \frac{17}{7}$. $Q(-\frac{13}{7}, \frac{17}{7})$. Then $k_1+k_2 = \frac{-13+17}{7} = \frac{4}{7}$.

We must have made an error in the given answer. Let's find the equation of AQ. $A(-3,3), Q(k_1, k_2)$. Since Q lies on BC: $3k_1 - k_2 + 8 = 0$, or $k_2 = 3k_1 + 8$. Since Q lies on AP: $k_1 + 2k_2 - 3 = 0$, or $k_1 + 2(3k_1 + 8) - 3 = 0$, or $k_1 + 6k_1 + 16 - 3 = 0$, so $7k_1 + 13 = 0$, or $k_1 = -\frac{13}{7}$. $k_2 = 3(-\frac{13}{7}) + 8 = -\frac{39}{7} + \frac{56}{7} = \frac{17}{7}$. $k_1 + k_2 = \frac{-13+17}{7} = \frac{4}{7}$. The given answer of 2 is incorrect.

Common Mistakes & Tips

• Double-check the arithmetic when solving simultaneous equations to avoid errors.
• Always verify the validity of the solution with the given conditions (e.g., $ab > 0$).
• When encountering a contradiction, carefully review each step, starting from the basic concepts.

Summary

We used the circumcentre property to find the coordinates of the vertices of the triangle. Then, we found the equations of the lines AP and BC and solved them simultaneously to find the intersection point Q. Finally, we calculated the sum of the coordinates of Q. The correct sum is $\frac{4}{7}$.

Final Answer

The final answer is \boxed{\frac{4}{7}}, which corresponds to option (B).