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

Question

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

Aicharya · Accepted Answer

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 \quad 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 $\triangle A B C=3 \times \triangle A G C$ $ =3 \times \frac{1}{2}[1(1-3)+4(3-2)+4(2-1)]=9 . $

In a triangle $A B C$, coordianates of $A$ are $(1,2)$ and the equations of the medians through $B$ and $C$…

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