How to find the area of triangle?

Question

$A =(3, 1)$ 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.

Answer 1

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}

Answer 2

You have a formula with determinants: $$\operatorname{area}(OAA')=\frac12\Bigl|\det\biggl(\overrightarrow{OA},\overrightarrow{OA'}\Bigr)\biggr|.$$

Answer 3

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 $AA'$.

Answer 4

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 $$A=\sqrt{(\sqrt{10}+\sqrt{5})((\sqrt{10}+\sqrt{5})-\sqrt{10})((\sqrt{10}+\sqrt{5})-\sqrt{10})((\sqrt{10}+\sqrt{5})-2\sqrt{5})}$$ $$=\sqrt{(\sqrt{10}+\sqrt{5})(\sqrt{10}+\sqrt{5}-\sqrt{10})(\sqrt{10}+\sqrt{5}-\sqrt{10})(\sqrt{10}+\sqrt{5}-2\sqrt{5})}$$ $$=\sqrt{(\sqrt{10}+\sqrt{5})(\sqrt{5})(\sqrt{5})(\sqrt{10}-\sqrt{5})}=\sqrt{5(\sqrt{10}+\sqrt{5})(\sqrt{10}-\sqrt{5})}$$ $$=\sqrt{5(10-\sqrt{50}+\sqrt{50}-5)}=\sqrt{5(5)}=5$$