Question

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

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

Answer 1

The Correct Option is D

The centroid \( (\alpha, \beta) \) of a triangle with vertices \( A(x_1, y_1), B(x_2, y_2), C(x_3, y_3) \) is given by: \[ \alpha = \frac{x_1 + x_2 + x_3}{3}, \quad \beta = \frac{y_1 + y_2 + y_3}{3}. \] Here, the vertices are \( A(6, 8), B(5, -7), C(1, 2) \).
The centroid is calculated as: \[ \alpha = \frac{6 + 5 + 1}{3} = \frac{12}{3} = 4, \quad \beta = \frac{8 + (-7) + 2}{3} = \frac{3}{3} = 1. \] Now, the given relations are: \[ \alpha + 2\beta = 4 + 2(1) = 6, \quad 2\alpha - \beta = 2(4) - 1 = 7. \] Thus, the roots of the quadratic equation are \( \alpha + 2\beta = 6 \) and \( 2\alpha - \beta = 7 \).
The quadratic equation with these roots is: \[ x^2 - (\alpha + 2\beta + 2\alpha - \beta)x + (\alpha + 2\beta)(2\alpha - \beta) = 0 \] Substitute the values: \[ x^2 - 13x + 42 = 0. \]