$A (3,1)$ 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?
 A: If $A'(h,k)$ is reflected point
the midpoint
$$\left(\dfrac{h+3}2,\dfrac{k+1}2\right)$$ will lie on $y=2x$
and $y=2x$ will be perpendicular to $AA'$
$$2\cdot\dfrac{k-1}{h-3}=-1$$
Solve the two simultaneous equations for $h,k$
Use https://www.mathopenref.com/coordtrianglearea.html
A: 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 
$$A = \frac12 |OA|^2 = 5 $$
A: Exploit the symmetry in the problem.  
Let $B$ be the midpoint of $A$ and $A'$. By symmetry, $\triangle{AOB}\cong\triangle{A'OB}$, but we can just as well place a copy of $\triangle{AOB}$ along $\overline{AB}$ instead to form a paralellogram, as shown here:

Thus, the area of $\triangle{AOA'}$ is equal to the area of this paralellogram. This in turn is equal to the determinant of the matrix with vectors $\vec{OA}$ and $\vec{OB}$ as its rows or columns. (Or, if you prefer, the norm of the “cross product” of $\vec{OA}$ and $\vec{OB}$.)  
Now, $B$ is the orthogonal projection of $A$ onto the line of reflection. The latter has direction vector $(1,2)$, so using a standard formula for orthogonal projection we have $$\operatorname{area}(\triangle{A'OA}) = \det\begin{bmatrix}\vec{OA}\\\vec{OB}\end{bmatrix} = {(3,1)\cdot(1,2)\over(1,2)\cdot(1,2)}\det\begin{bmatrix}3&1 \\ 1&2\end{bmatrix} = 5.$$ In the next-to-last step, I’ve factored the coefficient of the direction vector $(1,2)$ that comes from the orthogonal projection formula out of the bottom row of the determinant to reduce clutter.  
Of course, one can also compute $B$ the way you did, by intersecting the line of reflection with its perpendicular through $A$. Either way, you don’t need to compute the reflected point $A'$ explicitly.
