In a triangle ABC, coordinates of A are (1,2) and the equations of the medians through B and C are respectively, $x+y=5$ and $x=4$. Then area of $\Delta ABC$ (in sq.units) is

We draw the triangle ABC by given data below

So, centroid G is (4,1)

Let coordinate of $B(\alpha ,5-\alpha )$ and $B(4,\beta )$.

Mid-point AB lies of median of $\overline{)C}$ and has value $M\left(\frac{{\displaystyle \alpha +1}}{{\displaystyle 2}},\frac{{\displaystyle 1-\alpha }}{{\displaystyle 2}}\right)$

So, $\frac{\alpha +1}{2}=4\Rightarrow \alpha =7$

Similarly, mis-point AC lies of median of $\overline{)B}$ and has value $N\left(\frac{{\displaystyle 5}}{{\displaystyle 2}},\frac{{\displaystyle \beta +2}}{{\displaystyle 2}}\right)$

 $\frac{5}{2}+\frac{\beta +2}{2}=5\Rightarrow \beta =3$

Therefore, area of $\Delta ABC$ is $\frac{{\displaystyle 1}}{{\displaystyle 2}}\left|1(-2-3)+7(3+2)+4(2+2)\right|=\frac{1}{2}\left|-5+7+16\right|=9$