Question

If $\alpha ,\beta$ and $\gamma$ are three different values, then equation
$\left(\beta -\gamma \right)x+\left(\gamma -\alpha \right)y+\left(\alpha -\beta \right)=0$ and $\left({\beta }^3-{\gamma }^3\right)x+\left({\gamma }^3-{\alpha }^3\right)y+\left({\alpha }^3-{\beta }^3\right)=0$
represents same line, if

• A. $\alpha +\beta +\gamma =0$
• B. $\alpha +\beta +\gamma =1$
• C. $\alpha =\beta +\gamma$
• D. None of these

Answer 1

The correct option is A $\alpha +\beta +\gamma =0$

If these two represent the same line.Hence,

$(\beta -\gamma )$ = ${\beta }^3-{\gamma }^3$

$\beta -\gamma =(\beta -\gamma {)}^3+3\beta \gamma (\beta -\gamma )$

$1=(\beta -\gamma {)}^2+3\beta \gamma$

$1={\beta }^2+{\gamma }^2+\beta \gamma$

Similarily,

$1={\alpha }^2+{\beta }^2+\alpha \beta$.................(i)

$1={\gamma }^2+{\beta }^2+\gamma \beta$............(ii)

$1={\alpha }^2+{\gamma }^2+\alpha \gamma$................(iii)

Subtracting (iii) from (ii),we get

$0={\alpha }^2-{\beta }^2+\gamma (\alpha -\beta )$

$0=(\alpha +\beta )(\alpha -\beta )+\gamma (\alpha -\beta )$

$0=(\alpha -\beta )(\alpha +\beta +\gamma )$

As, $\alpha ,\beta ,\gamma$ are all different value

Hence, $\alpha \ne \beta$ the only solution left is

$\alpha +\beta +\gamma =0$