Mathematics - Point Question with Solution | TestHub

Median through $C$ is $x=4$ So the $x$ coordianate of $C$ is 4 . let $C\equiv (4,y)$, then the midpoint of $A(1,2)$ and $C(4,y)$ is $D$ which lies on the median through $B$.
$$\therefore D\equiv \left(\frac{1+4}{2},\frac{2+y}{2}\right)$$ Now $\frac{1+4+2+y}{2}=5\Rightarrow y=3$. So, $C\equiv (4,3)$. The centroid of the triangle is the intersection of the mesians. Here the medians $x=4$ and $x+4$ and $x+y=5$ intersect at $G(4,1)$. The area of triangle $△ABC=3\times △AGC$ $$=3\times \frac{1}{2}[1(1-3)+4(3-2)+4(2-1)]=9.$$