Let A ( a ,   0 ) ,   B ( b ,   2 b + 1 ) and C ( 0 , b ) ,   b ≠ 0 ,   | b |…

Given coordinates of triangle $ABC$, $\mathrm{A}(\mathrm{a},0),\mathrm{B}(\mathrm{b},2\mathrm{b}+1),\mathrm{C}(0,\mathrm{b})$

Area$=\frac{1}{2}\left|\begin{array}{ccc}a& 0& 1\\ b& 2b+1& 1\\ 0& b& 1\end{array}\right|$

$\frac{1}{2}\left[\mathrm{a}(\mathrm{b}+1)+{\mathrm{b}}^2\right]=\pm 1$

$ab+a+b^2=\pm 2$

If  $2$ is positive 

$a=\frac{2-b^2}{b+1}$

If  $2$is negative

$a=\frac{-2-b^2}{b+1}$

Sum of possible values of $a=\frac{2-b^2}{b+1}+\frac{-2-b^2}{b+1}$

$=\frac{-2b^2}{b+1}$