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 $$= 1$ with its vertex at $(1,1)$. I tried to solve the question but could not approach it properly. Can someone help?
Hint: find point $B(x,y)$ by solving the system:
$$ \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 $y$.