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The vertices of a triangle ABC are A(1,0) while B and C lie on the parabola ${ y = 2x - x^2}$. If AB = AC = $\frac{\sqrt{7}}{3}$ , then it's area in sq. units is.………

I plotted the parabola and it came to be symmetrical along the line $x $$= 1$ with its vertex at $(1,1)$. I tried to solve the question but could not approach it properly. Can someone help?

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    ?The question is to find the area? – 2017-01-08
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    Yes area is to be determined – 2017-01-08
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    I hope you won't mind. Please encode the image next time if you have similar question like this. This is for the benefit of all the readers. – 2017-01-08
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    Actually I don't know how to encode the image. It would be helpful if someone make the necessary edit. Thanks for you r suggestion. – 2017-01-08
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    You could follow [this guide](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) and do it yourself. – 2017-01-08
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    Try to edit and eventually you will learn. – 2017-01-08
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    @Resorcinol One thing more, your vertex must be $(1,1)$ and not $(1,-1)$. – 2017-01-08
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    @juniven yes, I wrote it in a hurry. Thanks! – 2017-01-08

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Hint: find point $B(x,y)$ by solving the system:

$$ \begin{cases} \begin{align} y & = 2x-x^2 \\ (x-1)^2 + y^2 & = \frac{7}{9} \end{align} \end{cases} $$

For a shortcut, note that the first equation can be rewritten as $y=-(x-1)^2+1$, then eliminating the $(x-1)^2$ term between the two equations gives a simple quadratic in $y$.