Let - - Y be the roots of the equation x 3-12 x 2-44 x -48-0- If the coordinates of the vertices of a triangle are A - 1- B - 1- and C Y - 1- Y - then the centroid of the ABC isA- 11-36- 4B- 4- 11-36C- 3- 5-11D- 5-11- 3

The correct option is B (4, 1136)Given : x^3 12x^2 + 44x 48 = 0 Centroid = (α + β + γ3, 1α+1β+1γ3) Now, α + β +γ3 = 123 = 4 13(1α+ 1β + 1γ) = 13(αβ + ...

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Standard XII

Mathematics

Centroid

Let α, β, Y b...

Question

Let $α, β, γ$ be the roots of the equation $x^{3} - 12 x^{2} + 44 x - 48 = 0.$ If the coordinates of the vertices of a triangle are $A (α, \frac{1}{α}),$ $B (β, \frac{1}{β})$ and $C (γ, \frac{1}{γ}),$ then the centroid of the $△ A B C$ is

(4, \frac{11}{36})

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(\frac{11}{36}, 4)

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(3, \frac{5}{11})

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(\frac{5}{11}, 3)

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Solution

The correct option is A $(4, \frac{11}{36})$
Given : $x^{3} - 12 x^{2} + 44 x - 48 = 0$

Centroid $= ⎛ ⎜ ⎜ ⎜ ⎝ \frac{α + β + γ}{3}, \frac{\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}}{3} ⎞ ⎟ ⎟ ⎟ ⎠$

Now, $\frac{α + β + γ}{3} = \frac{12}{3} = 4$

$\frac{1}{3} (\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}) = \frac{1}{3} (\frac{α β + β γ + γ α}{α β γ}) = \frac{1}{3} (\frac{44}{48}) = \frac{11}{36}$

Therefore, centroid $= (4, \frac{11}{36})$

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Centroid

Standard XII Mathematics

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Let α,β,γ be the roots of the equation x3−12x2+44x−48=0. If the coordinates of the vertices of a triangle are A(α,1α), B(β,1β) and C(γ,1γ), then the centroid of the △ABC is

The correct option is A (4,1136)Given : x3−12x2+44x−48=0 Centroid =⎛⎜ ⎜ ⎜⎝α+β+γ3,1α+1β+1γ3⎞⎟ ⎟ ⎟⎠ Now, α+β+γ3=123=4 13(1α+1β+1γ)=13(αβ+βγ+γααβγ)=13(4448)=1136 Therefore, centroid =(4,1136)

Let $α, β, γ$ be the roots of the equation $x^{3} - 12 x^{2} + 44 x - 48 = 0.$ If the coordinates of the vertices of a triangle are $A (α, \frac{1}{α}),$ $B (β, \frac{1}{β})$ and $C (γ, \frac{1}{γ}),$ then the centroid of the $△ A B C$ is