If $\begin{vmatrix}
{2a}&{x_1} &{y_1}\\
{2b}&{x_2}& {y_2} \\
{2c}&{x_3}&{y_3}\\
\end{vmatrix}$ $ = $$\frac {abc}{2} \neq 0$, then the area of the triangle whose vertices are $(\frac {x_1}{a},\frac {y_1}{a})$$(\frac {x_2}{b},\frac {y_2}{b})$$(\frac {x_3}{c},\frac {y_3}{c})$ is