Let a, b, c be in arithmetic progression. Let | vedclass

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(D) The centroid of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\righ

Vedclass · Accepted Answer

$-\frac{71}{256}$

Vedclass · Answer

(D) The centroid of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.Given vertices are $(a, c), (2, b), (a, b)$ and centroid is $\left(\frac{10}{3}, \frac{7}{3}\right)$.Equating the $x$-coordinates: $\frac{a+2+a}{3} = \frac{10}{3} \implies 2a + 2 = 10 \implies 2a = 8 \implies a = 4$.Equating the $y$-coordinates: $\frac{c+b+b}{3} = \frac{7}{3} \implies c + 2b = 7$.Since $a, b, c$ are in arithmetic progression,$2b = a + c$. Substituting $a=4$,we get $2b = 4 + c \implies c = 2b - 4$.Substitute $c$ into the equation $c + 2b = 7$: $(2b - 4) + 2b = 7 \implies 4b = 11 \implies b = \frac{11}{4}$.The quadratic equation is $4x^{2} + \frac{11}{4}x + 1 = 0$.For roots $\alpha, \beta$,we have $\alpha + \beta = -\frac{b}{a} = -\frac{11/4}{4} = -\frac{11}{16}$ and $\alpha\beta = \frac{c}{a} = \frac{1}{4}$.We need to find $\alpha^{2} + \beta^{2} - \alpha\beta = (\alpha + \beta)^{2} - 3\alpha\beta$.Substituting the values: $\left(-\frac{11}{16}\right)^{2} - 3\left(\frac{1}{4}\right) = \frac{121}{256} - \frac{3}{4} = \frac{121 - 192}{256} = -\frac{71}{256}$.

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