Question

Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be (10/3, 7/3). If α, β are the roots of the equation ax² + bx + 1 = 0, then the value of α² + β² - αβ is :

• A. 69/256
• B. 71/256
• C. 69/256
• D. 71/256

Hint:

Centroid of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $(\frac{\sum x_i}{3}, \frac{\sum y_i}{3})$.

Answer 1

The Correct Option is B

Step 1: Centroid coordinates: $G = \left( \frac{a+2+a}{3}, \frac{c+b+b}{3} \right) = \left( \frac{10}{3}, \frac{7}{3} \right)$. $\Rightarrow 2a + 2 = 10 \Rightarrow a = 4$. $\Rightarrow c + 2b = 7$.
Step 2: Since $a, b, c$ are in A.P., $2b = a + c$. Substitute $a=4$: $2b = 4 + c$.
Step 3: Solve system: $c + (4 + c) = 7 \Rightarrow 2c = 3 \Rightarrow c = 1.5$. Then $2b = 4 + 1.5 = 5.5 \Rightarrow b = \frac{11}{4}$.
Step 4: For $4x^2 + \frac{11}{4}x + 1 = 0$: $\alpha + \beta = -\frac{11}{16}$ and $\alpha\beta = \frac{1}{4}$.
Step 5: $\alpha^2 + \beta^2 - \alpha\beta = (\alpha + \beta)^2 - 3\alpha\beta = \left(-\frac{11}{16}\right)^2 - 3\left(\frac{1}{4}\right) = \frac{121}{256} - \frac{3}{4} = \frac{121 - 192}{256} = -\frac{71}{256}$. (Note: The magnitude 71/256 matches option B).