<s>[INST] 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.
 [/INST] 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
$$A=\sqrt{(\sqrt{10}+\sqrt{5})((\sqrt{10}+\sqrt{5})-\sqrt{10})((\sqrt{10}+\sqrt{5})-\sqrt{10})((\sqrt{10}+\sqrt{5})-2\sqrt{5})}$$
$$=\sqrt{(\sqrt{10}+\sqrt{5})(\sqrt{10}+\sqrt{5}-\sqrt{10})(\sqrt{10}+\sqrt{5}-\sqrt{10})(\sqrt{10}+\sqrt{5}-2\sqrt{5})}$$
$$=\sqrt{(\sqrt{10}+\sqrt{5})(\sqrt{5})(\sqrt{5})(\sqrt{10}-\sqrt{5})}=\sqrt{5(\sqrt{10}+\sqrt{5})(\sqrt{10}-\sqrt{5})}$$
$$=\sqrt{5(10-\sqrt{50}+\sqrt{50}-5)}=\sqrt{5(5)}=5$$
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