Key Concepts and Formulas

• Circumcenter: The point of intersection of the perpendicular bisectors of the sides of a triangle. It is equidistant from the vertices. For a right-angled triangle, the circumcenter is the midpoint of the hypotenuse.
• Slope of a line: For a line $Ax + By = C$, the slope $m = -\frac{A}{B}$.
• Perpendicular lines: Two lines with slopes $m_1$ and $m_2$ are perpendicular if $m_1 \cdot m_2 = -1$.
• Midpoint Formula: The midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.

Step-by-Step Solution

Step 1: Identify the given lines and check for perpendicularity.

We are given the following lines: $L_1: 4x + 3y = 69$ $L_2: -3x + 4y = 17$ $L_3: x + 7y = 61$

We need to check if any pair of these lines are perpendicular. We do this by finding their slopes and checking if the product of any two slopes is -1.

The slopes are: $m_1 = -\frac{4}{3}$ $m_2 = \frac{3}{4}$ $m_3 = -\frac{1}{7}$

Now, let's check the products of the slopes: $m_1 \cdot m_2 = -\frac{4}{3} \cdot \frac{3}{4} = -1$ Since $m_1 \cdot m_2 = -1$, the lines $L_1$ and $L_2$ are perpendicular. Therefore, the triangle formed by these three lines is a right-angled triangle.

Step 2: Find the vertices of the triangle.

Since we know the triangle is right-angled, we can find the vertices by finding the intersection points of the lines.

• Vertex A: Intersection of $L_1$ and $L_3$. We solve the system of equations: $4x + 3y = 69$ $x + 7y = 61$ From the second equation, $x = 61 - 7y$. Substituting into the first equation: $4(61 - 7y) + 3y = 69$ $244 - 28y + 3y = 69$ $-25y = -175$ $y = 7$ Substituting $y = 7$ back into $x = 61 - 7y$, we get $x = 61 - 7(7) = 61 - 49 = 12$. Thus, Vertex A is $(12, 7)$.

• Vertex B: Intersection of $L_2$ and $L_3$. We solve the system of equations: $-3x + 4y = 17$ $x + 7y = 61$ From the second equation, $x = 61 - 7y$. Substituting into the first equation: $-3(61 - 7y) + 4y = 17$ $-183 + 21y + 4y = 17$ $25y = 200$ $y = 8$ Substituting $y = 8$ back into $x = 61 - 7y$, we get $x = 61 - 7(8) = 61 - 56 = 5$. Thus, Vertex B is $(5, 8)$.

• Vertex C: Intersection of $L_1$ and $L_2$. This vertex corresponds to the right angle. We solve the system of equations: $4x + 3y = 69$ $-3x + 4y = 17$ Multiply the first equation by 3 and the second equation by 4 to eliminate $x$: $12x + 9y = 207$ $-12x + 16y = 68$ Adding the equations, we get $25y = 275$, so $y = 11$. Substituting $y = 11$ into the first equation: $4x + 3(11) = 69$, so $4x = 36$, and $x = 9$. Thus, Vertex C is $(9, 11)$.

Step 3: Determine the circumcenter $C(\alpha, \beta)$.

Since the triangle is right-angled at C, the hypotenuse is AB. The circumcenter is the midpoint of the hypotenuse. Using the midpoint formula with $A(12, 7)$ and $B(5, 8)$: $C(\alpha, \beta) = \left(\frac{12 + 5}{2}, \frac{7 + 8}{2}\right) = \left(\frac{17}{2}, \frac{15}{2}\right)$ Therefore, $\alpha = \frac{17}{2}$ and $\beta = \frac{15}{2}$.

Step 4: Evaluate the expression $(\alpha - \beta)^2 + \alpha + \beta$.

Substitute the values of $\alpha$ and $\beta$: $(\alpha - \beta)^2 + \alpha + \beta = \left(\frac{17}{2} - \frac{15}{2}\right)^2 + \frac{17}{2} + \frac{15}{2} = \left(\frac{2}{2}\right)^2 + \frac{32}{2} = (1)^2 + 16 = 1 + 16 = 17$

Common Mistakes & Tips

• Always check for perpendicular lines first. Recognizing a right-angled triangle significantly simplifies the problem.
• Be careful with signs when calculating slopes and substituting values.
• Double-check your calculations, especially when solving systems of equations.

Summary

We identified that the triangle formed by the given lines is a right-angled triangle. We then found the vertices of the triangle by finding the intersection points of the lines. Since the triangle is right-angled, the circumcenter is the midpoint of the hypotenuse. Finally, we calculated the coordinates of the circumcenter and substituted them into the given expression to find the answer.

Final Answer

The final answer is \boxed{17}, which corresponds to option (B).