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Straight Lines Ncert Solutions Maths class 11th Maths Class 11th

Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be $(\frac{10}{3}, \frac{7}{3})$ . If a, b are the roots of the equation $a x^{2} + b x + 1 = 0,$ then the value of $α^{2} + β^{2} - α β$ is:

Option 1 -

$- \frac{69}{256}$

Option 2 -

$\frac{69}{256}$

Option 3 -

$\frac{27}{256}$

Option 4 -

$- \frac{71}{256}$

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

$\frac{a + 2 + a}{3} = \frac{10}{3}$

2a + 2 = 0

2a = 8 -> a = 4 .(i)

and $\frac{c + b + b}{3} = \frac{7}{3}$

2b + c = 7 .(ii)

Since a, b, c are in A.P.

2b = a + c

From (i) $∴$ 2b = 4 + c .(iii)

Solving (ii) and (iii)

4 + c + c = 7

2c = 3

$c = \frac{3}{2}$

$∴ 2 b = 4 + \frac{3}{2} = \frac{11}{2}$

$∴ b = \frac{11}{4}$

As per question

$α + β = \frac{- b}{a} a n d α β = \frac{1}{a}$

$= \frac{121 - 192}{256} = \frac{- 71}{256}$

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Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be $(\frac{10}{3}, \frac{7}{3})$ . If a, b are the roots of the equation $a x^{2} + b x + 1 = 0,$ then the value of $α^{2} + β^{2} - α β$ is:

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Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be ( 1 0 3 , 7 3 ) . If a, b are the roots of the equation a x 2 + b x + 1 = 0 , then the value of α 2 + β 2 − α β is:

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Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be $(\frac{10}{3}, \frac{7}{3})$ . If a, b are the roots of the equation $a x^{2} + b x + 1 = 0,$ then the value of $α^{2} + β^{2} - α β$ is: