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In a triangle $ABC$, 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 $\Delta ABC$ (in sq. units) is
$5$
$9$
$12$
$4$
Solution
Median through $C$ is $x=4$
So the $x$ coordianate of $C$ is $4$. let $C \equiv \left( {4,y} \right)$
then the midpoint of $A(1,2)$ and $C(4,y)$ is $D$
which lies on the median through $B$.
$D \equiv \left( {\frac{{1 + 4}}{2},\frac{{2 + y}}{2}} \right)$
Now, $\frac{{1 + 4 + 2 + y}}{2} = 5 \Rightarrow y = 3$
So, $C \equiv \left( {4,3} \right)$.
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 $\Delta ABC = 3 \times \Delta AGC$
$ = 3 \times \frac{1}{2}\left[ {1\left( {1 – 3} \right) + 4\left( {3 – 2} \right) + 4\left( {2 – 1} \right)} \right] = 9$
