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Q.
Let $A ( a , 0), B ( b , 2 b +1)$ and $C (0, b ), b \neq 0,| b | \neq 1$, be points such that the area of triangle $ABC$ is $1 sq$. unit, then the sum of all possible values of a is :
$\left|\frac{1}{2} \begin{vmatrix} a & 0 & 1 \\b & 2 b +1 & 1 \\0 & b & 1\end{vmatrix} \right|=1 $
$\Rightarrow \left|\begin{vmatrix} a & 0 & 1 \\b & 2 b +1 & 1 \\0 & b & 1 \end{vmatrix}\right|=\pm 2 $
$\Rightarrow a (2 b +1- b )-0+1\left( b ^{2}-0\right)=\pm 2$
$\Rightarrow a =\frac{\pm 2- b ^{2}}{ b +1} $
$\therefore a =\frac{2- b ^{2}}{ b +1} $ and $ a =\frac{-2- b ^{2}}{ b +1}$
sum of possible values of '$a$' is
$=\frac{-2 b^{2}}{a+1}$