In a triangle ABC, coordianates of A are (1, | vedclass

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Question

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(

Vedclass · Accepted Answer

$9$

Vedclass · Answer

Median through $C$ is $x=4$

So the $x$ coordianate of $C$ is $4$. let $C \equiv \left( {4,y} ight)$

then the midpoint of $A(1,2)$ and $C(4,y)$ is $D$ 

which lies on the median through $B$.

$D \equiv \left( {\frac{{1 + 4}}{2},\frac{{2 + y}}{2}} ight)$

Now, $\frac{{1 + 4 + 2 + y}}{2} = 5 \Rightarrow y = 3$

So, $C \equiv \left( {4,3} ight)$.

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 $\Delta ABC = 3 	imes \Delta AGC$

$ = 3 	imes \frac{1}{2}\left[ {1\left( {1 - 3} ight) + 4\left( {3 - 2} ight) + 4\left( {2 - 1} ight)} ight] = 9$

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

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