If $m_1$ and $m_2$ are the roots of the equation $x^2-ax-a-1=0$ , then the area of the triangle  formed by the three straight lines $y=m_{1}x,y=m_{2}x and y=a(a\ne -1)$is

Since  $m_1$ and $m_2$ are the roots of the equation $x^2-ax-a-1=0$

so $m_1+m_2=a;m_1{m}_2=-(a+1)$

$\Rightarrow {(m_1-m_2)}^2={(m_1+m_2)}^2+4m_1{m}_2$

$=a^2+4(a+1)$$={(a+2)}^2$

$\therefore$ The required area is $\Delta =\pm \frac{a^2(a+2)}{2(a+1)}$

since area is positive quantity, so we get two answers