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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 $\triangle A B C$ (in sq. units) is
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The correct answer is:
9
9
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 $\triangle A B C=3 \times \triangle A G C$
$$
=3 \times \frac{1}{2}[1(1-3)+4(3-2)+4(2-1)]=9 .
$$
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 $\triangle A B C=3 \times \triangle A G C$
$$
=3 \times \frac{1}{2}[1(1-3)+4(3-2)+4(2-1)]=9 .
$$
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