Question

Let α,β,γ be the roots of the equation 2x³+3x²-12x+3= 0 and A(α,β,γ),B(β,γ,α),C(γ,α,β) represent vertices of a triangle ABC then the centroid of the triangle lies upon the line.

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• α,β,γ be the roots of the equation 2x³+3x²-12x+3= 0.

• A(α,β,γ),B(β,γ,α),C(γ,α,β) represent vertices of a triangle ABC.

We know,

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: roots \: of \: a {x}^{3} + b {x}^{2} + cx + d = 0, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

So, given that,

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: roots \: of \:2{x}^{3} +3{x}^{2} - 12x + 3 = 0, \: then}$

$\rm :\longmapsto\:\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{3}{2}}} - - - - (1)$

Now, further, It is given that,

• A(α,β,γ),B(β,γ,α),C(γ,α,β) represent vertices of a triangle ABC.

So, Let assume centroid of triangle ABC, is represented as G(x, y, z) and is given by

$\rm :\longmapsto\:(x,y,z) = \bigg(\dfrac{ \alpha + \beta + \gamma }{3},\dfrac{ \alpha + \beta + \gamma }{3},\dfrac{ \alpha + \beta + \gamma }{3} \bigg)$

can be rewritten as

$\rm :\longmapsto\:(x,y,z) = \bigg( - \dfrac{1}{2}, - \dfrac{1}{2}, - \dfrac{1}{2} \bigg)$

$\red{ \bigg\{ \sf \: \because \: using \: equation \: (1) \bigg\}}$

So,

$\rm\implies \:x \: = \: - \: \dfrac{1}{2}$

$\rm\implies \:y \: = \: - \: \dfrac{1}{2}$

$\rm\implies \:z \: = \: - \: \dfrac{1}{2}$

So,

$\bf\implies \:x = y = z$

It means coordinates of Centroid G, lies on the above line which passes through (0, 0, 0) and having direction ratios (1, 1, 1) respectively.

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

α,β,γ be the roots of the equation 2x³+3x²-12x+3= 0.

A(α,β,γ),B(β,γ,α),C(γ,α,β) represent vertices of a triangle ABC.

We know,

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: roots \: of \: a {x}^{3} + b {x}^{2} + cx + d = 0, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

So, given that,

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: roots \: of \:2{x}^{3} +3{x}^{2} - 12x + 3 = 0, \: then}$

$\rm :\longmapsto\:\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{3}{2}}} - - - - (1)$

Now, further, It is given that,

A(α,β,γ),B(β,γ,α),C(γ,α,β) represent vertices of a triangle ABC.

So, Let assume centroid of triangle ABC, is represented as G(x, y, z) and is given by

$\rm :\longmapsto\:(x,y,z) = \bigg(\dfrac{ \alpha + \beta + \gamma }{3},\dfrac{ \alpha + \beta + \gamma }{3},\dfrac{ \alpha + \beta + \gamma }{3} \bigg)$

can be rewritten as

$\rm :\longmapsto\:(x,y,z) = \bigg( - \dfrac{1}{2}, - \dfrac{1}{2}, - \dfrac{1}{2} \bigg)$

$\red{ \bigg\{ \sf \: \because \: using \: equation \: (1) \bigg\}}$

So,

$\rm\implies \:x \: = \: - \: \dfrac{1}{2}$

$\rm\implies \:y \: = \: - \: \dfrac{1}{2}$

$\rm\implies \:z \: = \: - \: \dfrac{1}{2}$

So,

$\bf\implies \:x = y = z$

It means coordinates of Centroid G, lies on the above line which passes through (0, 0, 0) and having direction ratios (1, 1, 1) respectively.