Key Concepts and Formulas

• Euler Line Property: For any triangle, its circumcentre ($C$), centroid ($G$), and orthocentre ($H$) are collinear, and $G$ divides the line segment $CH$ in the ratio $1:2$, i.e., $\vec{OG} = \frac{\vec{OH} + 2\vec{OC}}{3}$.
• Orthocentre of a Triangle: The orthocentre is the point of intersection of the altitudes of a triangle. An altitude is a line segment from a vertex perpendicular to the opposite side.
• Perpendicular Lines: If two lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1 m_2 = -1$. The slope of a line $Ax+By+C=0$ is given by $-\frac{A}{B}$.

Step-by-Step Solution

Step 1: Find the Centroid of the Second Triangle We are given the circumcentre $C_2 = (3,4)$ and orthocentre $H_2 = (-6,-8)$ of the second triangle. We need to find the coordinates of its centroid $G_2$. Using the Euler line property, we have: $G_2 = \left( \frac{x_{H_2} + 2x_{C_2}}{3}, \frac{y_{H_2} + 2y_{C_2}}{3} \right)$ Substituting the given coordinates: $G_2 = \left( \frac{-6 + 2(3)}{3}, \frac{-8 + 2(4)}{3} \right) = \left( \frac{-6+6}{3}, \frac{-8+8}{3} \right) = (0,0)$ Thus, the centroid of the second triangle is $G_2 = (0,0)$. The problem states that this centroid is the orthocentre of the first triangle, so $H_1 = (0,0)$.

Step 2: Find the Intersection Point of the First Two Lines The first triangle is formed by the lines $L_1: 2x+3y-1=0$, $L_2: x+2y-1=0$, and $L_3: ax+by-1=0$. Let's find the intersection point $A$ of $L_1$ and $L_2$. From $L_2$, we have $x = 1-2y$. Substituting this into $L_1$: $2(1-2y) + 3y - 1 = 0$ $2 - 4y + 3y - 1 = 0$ $1 - y = 0 \implies y = 1$ Substituting $y=1$ back into $x = 1-2y$: $x = 1 - 2(1) = -1$ So, vertex $A = (-1,1)$.

Step 3: Use the Orthocentre Property for Altitude from A Since $H_1 = (0,0)$ is the orthocentre, the line segment $AH_1$ is perpendicular to the side $L_3$ (defined by $ax+by-1=0$). The slope of $AH_1$ is $m_{AH_1} = \frac{1-0}{-1-0} = -1$. The slope of $L_3$ is $m_{L_3} = -\frac{a}{b}$. Since $AH_1 \perp L_3$, we have $m_{AH_1} \cdot m_{L_3} = -1$, which implies: $(-1)\left(-\frac{a}{b}\right) = -1$ $\frac{a}{b} = -1 \implies a = -b$

Step 4: Find the Intersection Point of the Second and Third Lines Let $C = (x_C, y_C)$ be the intersection of $L_2: x+2y-1=0$ and $L_3: ax+by-1=0$. Since $H_1 = (0,0)$ is the orthocentre, the line segment $CH_1$ is perpendicular to $L_1: 2x+3y-1=0$. The slope of $L_1$ is $m_{L_1} = -\frac{2}{3}$. Thus, the slope of $CH_1$ is $m_{CH_1} = \frac{3}{2}$. Since $H_1=(0,0)$, we have $m_{CH_1} = \frac{y_C-0}{x_C-0} = \frac{y_C}{x_C} = \frac{3}{2}$. So, $y_C = \frac{3}{2}x_C$. Substituting this into the equation of $L_2$: $x_C + 2\left(\frac{3}{2}x_C\right) - 1 = 0$ $x_C + 3x_C - 1 = 0$ $4x_C = 1 \implies x_C = \frac{1}{4}$ Then $y_C = \frac{3}{2}\left(\frac{1}{4}\right) = \frac{3}{8}$. So, $C = \left(\frac{1}{4}, \frac{3}{8}\right)$.

Step 5: Use the Fact that Point C Lies on Line $L_3$ Since $C$ lies on $L_3: ax+by-1=0$, we substitute the coordinates of $C$ into the equation: $a\left(\frac{1}{4}\right) + b\left(\frac{3}{8}\right) - 1 = 0$ Multiplying by 8, we get: $2a + 3b - 8 = 0$

Step 6: Solve for a and b We have the system of equations:

1. $a = -b$
2. $2a + 3b - 8 = 0$ Substitute $a = -b$ into the second equation: $2(-b) + 3b - 8 = 0$ $-2b + 3b - 8 = 0$ $b = 8$ Then $a = -b = -8$.

Step 7: Calculate |a-b| We have $a = -8$ and $b = 8$. Then $|a-b| = |-8 - 8| = |-16| = 16$

Common Mistakes & Tips

• Ensure the correct ratio for the Euler line property is used.
• Carefully calculate slopes and apply the perpendicularity condition.
• Pay attention to signs when substituting values and solving equations.

Summary We used the Euler line property to find the centroid of the second triangle, which was also the orthocentre of the first triangle. Using the orthocentre property and the equations of the lines, we found the values of $a$ and $b$, and finally, we calculated $|a-b|$.

The final answer is \boxed{16}.