How to find the area of triangle? $A =(3, 1)$ is reflected over line $y = 2x$, become $A'$.
$O = (0,0)$
Find area of triangle OAA'
I find that $A'$ is $(-1, 3)$
I find it by mutiply $A$ to matrix of $y = 2x$
There is rule Area = $$\sqrt{s(s-a)(s-b)(s-c)}$$
$S = \frac{a+b+c}{2}$
$a$ = distance between $A$ and $O=\sqrt {10}$
$b$ = between $A'$ and $O = \sqrt {10}$
$c$ = between $A$ and $A' = \sqrt {20}$.
Am i on the right way? Because it looks so complicated to solve. I am afraid i am on the wrong way.
 A: You have a formula with determinants:
$$\operatorname{area}(OAA')=\frac12\Bigl|\det\biggl(\overrightarrow{OA},\overrightarrow{OA'}\Bigr)\biggr|.$$
A: You can be smarter than that.
When you draw line segment $AA'$ it intersects the mirror line $y=2x$ at $B$, and the segment is perpendicular to the mirror at that point.  So the area is just half the length of $AA'$ (base of the isosceles triangle) times the length of $OB$ (altitude).
Knowing that $A'=(-1,3)$, calculate the midpoint $B$ as $(1,2)$, then you readily get the length of $OB$.  Proceed from there with the length you already got for $AA'$.
A: 
Since $A_1=(-1,3)$, it follows that 
$\triangle A_1OA$ is the right triangle,
which is also isosceles,
hence its area
\begin{align}
S_{\triangle A_1OA}
&=
\tfrac12|OA||OA_1|
=\tfrac12|OA|^2
=\tfrac12(A_x^2+A_y^2)=5
. 
\end{align}  
Edit
Since you know the coordinates,
you can also use the coordinate-based formula
for the area:
\begin{align}
S_{\triangle A_1OA}
&=
\tfrac12\,\left|((A_x-O_x)\cdot(A_{1y}-O_y)-(A_{1x}-O_x)\cdot(A_y-O_y))\right|
\\
&=
\tfrac12\,|(A_x\cdot A_{1y}-A_{1x}\cdot A_y|
=\tfrac12\,|3\cdot3-(-1)\cdot 1|
=5
.
\end{align}
A: The long way:
The formula for the area of a  triangle with sides $a$, $b$ and $c$ is given by
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
where
$$s=\frac{a+b+c}{2}$$
With $a=\sqrt{10}$, $b=\sqrt{10}$ and $c=\sqrt{20}=2\sqrt{5}$,
$$s=\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}=\frac{2\sqrt{10}+2\sqrt{5}}{2}=\sqrt{10}+\sqrt{5}$$
The area is then
$$A=\sqrt{(\sqrt{10}+\sqrt{5})((\sqrt{10}+\sqrt{5})-\sqrt{10})((\sqrt{10}+\sqrt{5})-\sqrt{10})((\sqrt{10}+\sqrt{5})-2\sqrt{5})}$$
$$=\sqrt{(\sqrt{10}+\sqrt{5})(\sqrt{10}+\sqrt{5}-\sqrt{10})(\sqrt{10}+\sqrt{5}-\sqrt{10})(\sqrt{10}+\sqrt{5}-2\sqrt{5})}$$
$$=\sqrt{(\sqrt{10}+\sqrt{5})(\sqrt{5})(\sqrt{5})(\sqrt{10}-\sqrt{5})}=\sqrt{5(\sqrt{10}+\sqrt{5})(\sqrt{10}-\sqrt{5})}$$
$$=\sqrt{5(10-\sqrt{50}+\sqrt{50}-5)}=\sqrt{5(5)}=5$$
