$A (3,1)$ reflected to $y=2x$ what is area of AOA'? $y_1 = 2x$ = line of reflection
$x_1,y_1 = A(3,1) =$ reflected point
General formation of a line
$y = mx + k$
$Ax + By + C = 0$
Finding a line perpendicular to line of reflection
$m_2 = \frac{-1}{2}$
$y_2 = \frac{-1}{2}x + \frac{5}{2}$
Intersection of reflection line and line perpendicular to it
$y_1$ dan $y_2$ 
$4x = -x + 5$
$x = 1, y = 2$
Finding A'
Gradient from A' and 1,2 also -1/2
$\frac{y_3 - 2}{x_3 - 1} = \frac{-1}{2}$
$-2y_3 + 3 = x_3$
Distance 1,2 and 3,1 = $\sqrt 5$ 
Distance from 1,2 to A' also $\sqrt 5$
${(x_3 - 1)}^2+ {(y_3 - 2)}^2 = 5$
${(-2y_3 + 3)}^2+ {(y_3 - 2)}^2 = 5$
After find A' then find |OA'| |OA| and finf area. But is there less complicated way to find area? Am i missing something?

Note that $|OA| = \sqrt{3^2+1}=\sqrt{10}$ and the angle $\theta$ between $y=2x$ and OA is given by
$$\tan\theta = \tan(\theta_2-\theta_1) 
= \frac{\tan\theta_2 - \tan\theta_1}{1+\tan\theta_2\cdot \tan\theta_2  }
=\frac{2-\frac13}{1+2\cdot\frac13} = 1$$
which yields $\theta = 45^\circ$. Then, AOA' is an isosceles right triangle whose area is 
$$A = \frac12 |OA|^2 = 5 $$
