Key Concepts and Formulas

• Intersection of Lines: The point of intersection of two lines $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ can be found by solving the system of equations.
• Centroid of a Triangle: The centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
• Quadratic Equation from Roots: A quadratic equation with roots $r_1$ and $r_2$ can be written as $x^2 - (r_1 + r_2)x + r_1r_2 = 0$.

Step-by-Step Solution

Let the three given lines be: (L1) $15x - y = 82$ (L2) $6x - 5y = -4$ (L3) $9x + 4y = 17$

Step 1: Finding Vertex A (Intersection of L1 and L2)

We need to solve the system of equations:

1. $15x - y = 82$
2. $6x - 5y = -4$

• Strategy: Multiply equation (1) by 5 to eliminate $y$.
• Multiply (1) by 5: $5(15x - y) = 5(82) \implies 75x - 5y = 410 \quad \text{(Equation 1')}$
• Subtract equation (2) from equation (1'): $(75x - 5y) - (6x - 5y) = 410 - (-4)$ $69x = 414$
• Solve for $x$: $x = \frac{414}{69} = 6$
• Substitute $x = 6$ into equation (1): $15(6) - y = 82$ $90 - y = 82$ $y = 90 - 82 = 8$
• Vertex A is $(6, 8)$.

Step 2: Finding Vertex B (Intersection of L1 and L3)

We need to solve the system of equations:

1. $15x - y = 82$
2. $9x + 4y = 17$

• Strategy: Multiply equation (1) by 4 to eliminate $y$.
• Multiply (1) by 4: $4(15x - y) = 4(82) \implies 60x - 4y = 328 \quad \text{(Equation 1'')}$
• Add equation (3) to equation (1''): $(60x - 4y) + (9x + 4y) = 328 + 17$ $69x = 345$
• Solve for $x$: $x = \frac{345}{69} = 5$
• Substitute $x = 5$ into equation (3): $9(5) + 4y = 17$ $45 + 4y = 17$ $4y = 17 - 45 = -28$ $y = \frac{-28}{4} = -7$
• Vertex B is $(5, -7)$.

Step 3: Finding Vertex C (Intersection of L2 and L3)

We need to solve the system of equations: 2) $6x - 5y = -4$ 3) $9x + 4y = 17$

• Strategy: Multiply equation (2) by 4 and equation (3) by 5 to eliminate $y$.
• Multiply (2) by 4: $4(6x - 5y) = 4(-4) \implies 24x - 20y = -16 \quad \text{(Equation 2')}$
• Multiply (3) by 5: $5(9x + 4y) = 5(17) \implies 45x + 20y = 85 \quad \text{(Equation 3')}$
• Add equation (2') and equation (3'): $(24x - 20y) + (45x + 20y) = -16 + 85$ $69x = 69$
• Solve for $x$: $x = \frac{69}{69} = 1$
• Substitute $x = 1$ into equation (2): $6(1) - 5y = -4$ $6 - 5y = -4$ $-5y = -10$ $y = \frac{-10}{-5} = 2$
• Vertex C is $(1, 2)$.

Step 4: Calculating the Centroid $(\alpha, \beta)$

The vertices are $A(6, 8)$, $B(5, -7)$, and $C(1, 2)$.

• $\alpha = \frac{6 + 5 + 1}{3} = \frac{12}{3} = 4$
• $\beta = \frac{8 + (-7) + 2}{3} = \frac{3}{3} = 1$
• The centroid is $(\alpha, \beta) = (4, 1)$.

Step 5: Determining the Specific Roots

We need to find $\alpha + 2\beta$ and $2\alpha - \beta$.

• $r_1 = \alpha + 2\beta = 4 + 2(1) = 4 + 2 = 6$
• $r_2 = 2\alpha - \beta = 2(4) - 1 = 8 - 1 = 7$
• The roots are 6 and 7.

Step 6: Forming the Quadratic Equation

The quadratic equation is $x^2 - (r_1 + r_2)x + r_1r_2 = 0$.

• $r_1 + r_2 = 6 + 7 = 13$
• $r_1r_2 = 6 \times 7 = 42$
• The quadratic equation is $x^2 - 13x + 42 = 0$.

Common Mistakes & Tips

• Double-check arithmetic when solving systems of equations, especially with negative numbers.
• Ensure the centroid formula is applied correctly (sum of coordinates divided by 3).
• Carefully calculate the sum and product of the roots for the quadratic equation.

Summary

We found the vertices of the triangle by solving pairs of linear equations, then calculated the centroid using the vertex coordinates. Finally, we used the centroid's coordinates to find the two specific roots and formed the quadratic equation. The equation is $x^2 - 13x + 42 = 0$.

The final answer is $\boxed{x^{2}-13 x+42=0}$, which corresponds to option (B).