Let `(alpha,beta )` be the centroid of the triangle formed by the lines `15x - y= 82, 6x - 5y +(-4)` and `9x + 4y = 17 `. Then `alpha +2beta ` and `2alpha -beta ` are the roots of the equation

Options

A. `x^2-13x + 42 =0`
B. `x^2 - 7x +12 =0`
C. `x^2 - 10x +25 =0`
D. `x^2 -14x + 48 =0`

Answer

To solve the problem, we need to find the centroid of the triangle formed by the given lines and then use that information to derive the required equation.

### Step 1: Find the intersection points of the lines

We have three lines:
1. \( 15x - y = 82 \) (Line 1)
2. \( 6x - 5y = -4 \) (Line 2)
3. \( 9x + 4y = 17 \) (Line 3)

We will find the intersection points of these lines pairwise to get the vertices of the triangle.

**Intersection of Line 1 and Line 2:**
From Line 1, we can express \( y \):
\[ y = 15x - 82 \]

Substituting into Line 2:
\[ 6x - 5(15x - 82) = -4 \]
\[ 6x - 75x + 410 = -4 \]
\[ -69x + 410 = -4 \]
\[ -69x = -414 \]
\[ x = 6 \]

Now substituting \( x = 6 \) back into Line 1 to find \( y \):
\[ y = 15(6) - 82 = 90 - 82 = 8 \]

So, the first vertex \( A(6, 8) \).

**Intersection of Line 2 and Line 3:**
From Line 2, we can express \( y \):
\[ 5y = 6x + 4 \]
\[ y = \frac{6x + 4}{5} \]

Substituting into Line 3:
\[ 9x + 4\left(\frac{6x + 4}{5}\right) = 17 \]
\[ 9x + \frac{24x + 16}{5} = 17 \]
Multiplying through by 5 to eliminate the fraction:
\[ 45x + 24x + 16 = 85 \]
\[ 69x = 69 \]
\[ x = 1 \]

Now substituting \( x = 1 \) back into Line 2 to find \( y \):
\[ 6(1) - 5y = -4 \]
\[ 6 - 5y = -4 \]
\[ -5y = -10 \]
\[ y = 2 \]

So, the second vertex \( B(1, 2) \).

**Intersection of Line 1 and Line 3:**
From Line 1, we can express \( y \):
\[ y = 15x - 82 \]

Substituting into Line 3:
\[ 9x + 4(15x - 82) = 17 \]
\[ 9x + 60x - 328 = 17 \]
\[ 69x - 328 = 17 \]
\[ 69x = 345 \]
\[ x = 5 \]

Now substituting \( x = 5 \) back into Line 1 to find \( y \):
\[ y = 15(5) - 82 = 75 - 82 = -7 \]

So, the third vertex \( C(5, -7) \).

### Step 2: Calculate the centroid

The centroid \( ( \alpha, \beta ) \) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \) is given by:
\[ \alpha = \frac{x_1 + x_2 + x_3}{3}, \quad \beta = \frac{y_1 + y_2 + y_3}{3} \]

Substituting the vertices:
\[ \alpha = \frac{6 + 1 + 5}{3} = \frac{12}{3} = 4 \]
\[ \beta = \frac{8 + 2 - 7}{3} = \frac{3}{3} = 1 \]

### Step 3: Find \( \alpha + 2\beta \) and \( 2\alpha - \beta \)

Calculating:
\[ \alpha + 2\beta = 4 + 2(1) = 4 + 2 = 6 \]
\[ 2\alpha - \beta = 2(4) - 1 = 8 - 1 = 7 \]

### Step 4: Form the quadratic equation

The roots of the equation are \( 6 \) and \( 7 \). The quadratic equation with roots \( r_1 \) and \( r_2 \) can be formed as:
\[ x^2 - (r_1 + r_2)x + r_1 r_2 = 0 \]
Substituting the values:
\[ r_1 + r_2 = 6 + 7 = 13 \]
\[ r_1 r_2 = 6 \times 7 = 42 \]

Thus, the equation is:
\[ x^2 - 13x + 42 = 0 \]

### Final Answer
The required quadratic equation is:
\[ x^2 - 13x + 42 = 0 \]