Given that A, B and C do not lie on the same line...find the area of triangle ABC.

Question

\begin{array} { c } { \text { Given that } A , B \text { and } C \text { do not lie on the same line.If } \vec { O A } + \vec { O B } + \vec { O C } = 0 , | \vec { O A } | = \sqrt { 6 } \text { , } } \\ { | \vec { O B } | = 2 \text { and } | \vec { O C } | = \sqrt { 14 } , \text { find the area of } \triangle A B C . } \end{array}

I found this problem online and I am not sure what should I do. I have tried some geometric transformations but couldn't find the answer. I guess it isn't as easy as I thought at first.

Answer 1

Given condition can be rewriten like this $$O = {1\over 3}(A+B+C)=G$$so $O$ is actualy gravity center for $ABC$. If we reflect $G$ across the midpoint of $BC$ we get $G'$. Given lenghts are side lenghts of a triangle $BG'G$ which is 3 times smaller then $ABC$ and you are done.

Answer 2

Wlog $B=(2,0)$, $A=(x,y)$ with $x^2+y^2=6$, $C=(u,v)$ with $u^2+v^2=14$. Moreover, $x+u+2=0$ and $y+v=0$. This allows us to eliminate $u,v$: $$ \begin{align}x^2+y^2&=6\\(x+2)^2+y^2&=14\\\implies\qquad 4x+4=14-6\end{align}$$ and thereby $$ x=1,\;u=-3,\;y=\pm\sqrt 5,v=\mp\sqrt 5.$$ From $y+v=0$, we conclude that $BC$ intersects the $x$-axis at $\frac{x+u}{2}=-1$. Thus the $x$-axis splits $ABC$ into two triangles of base length $3$ and height $\sqrt 5$. The area is therefore $3\sqrt5$.