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In $\triangle \mathrm{ABC}$, if $\mathrm{A}=(1,2)$ and the equations of the medians through $B$ and $C$ are $x+y=5$ and $x=4$ respectively, then the area of $\triangle \mathrm{ABC}$ is
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The correct answer is:
$9$
$\mathrm{OC} \& \mathrm{OB}$ the equations of medians respectively and point $\mathrm{O}$ is the centroid.
$\because$ Intersection OC \& OB is centroid.
Solving $x+y=5 \& x=4$, we get the centroid as $(4,1)$.
Let the coordinate of point $C=(4, a)$
and coordinate of point $B=(b, 5-b)$
So, $\frac{1+4+b}{3}=4 \Rightarrow b=7$
$\frac{2+a+5-b}{3}=1 \Rightarrow a=3$
$\therefore A=(1,2) ; B=(7,-2) ; C=(4,3)$
$\Delta=\frac{1}{2}\left|\begin{array}{ccc}1 & 1 & 2 \\ 1 & 7 & -2 \\ 1 & 4 & 3\end{array}\right|$
$\Delta=\frac{1}{2} \times 18=9$
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