If the area of $ABC$ is $6$, what is the area of $A'B'C'$? In a triangle $ABC, AB = 3, BC = 4, CA = 5,$ Now $A$ is reflected at $B$ to $A'$ and $B$ is reflected 
at $C$ to $B'$ and  $C$ is reflected at $A$ to $C'$. Now if the area of $\triangle (ABC) = 6$ square unit, then area of $\triangle (A'B'C')$ is ... ?
Any hint is appreciated.
 A: It's worth noting that all we need to know to answer this question is the area of $\triangle ABC$, not its side lengths or vertex coordinates or any other information!
To see this, start from the determinant formula for the area of a triangle: $$S_{ABC} = \frac12 \left| \det \begin{bmatrix} x_A & y_A & 1 \\ x_B & y_B & 1 \\ x_C & y_C & 1\end{bmatrix} \right|$$ where $(x_A,y_A)$, $(x_B,y_B)$, $(x_C,y_C)$ are the coordinates of the three vertices.
The reflections that replace $A$ by $A'$, $B$ by $B'$, and $C$ by $C'$ can be written as $A' = 2B-A$, $B' = 2C-B$, and $C' = 2A-C$ in vector arithmetic, which corresponds to left-multiplying the matrix in the formula above by the matrix $$\begin{bmatrix}-1 & 2 & 0 \\ 0 & -1 & 2 \\ 2 & 0 & -1\end{bmatrix}.$$ So the effect on the area is to scale it by (the absolute value of) the determinant $$\det \begin{bmatrix}-1 & 2 & 0 \\ 0 & -1 & 2 \\ 2 & 0 & -1\end{bmatrix}=7.$$
Since the area of $\triangle ABC$ was $6$, the area of $\triangle A'B'C'$ is $7\cdot6 = 42$.
A: Following WW1 suggestion you may embed $ABC$ in $\mathbb{R}^2$ by assuming $A=(0,0), B=(3,0), C=(0,4)$ and getting $A'=(6,0), B'=(-3,8), C'=(0,-4)$ as a straightforward consequence. Then, by the shoelace formula,
$$ [A'B'C']=\frac{1}{2}\left|6\cdot 8+(-3)\cdot(-4)+0\cdot 0-0\cdot(-3)-8\cdot 0-(-4)\cdot 6\right| $$
i.e.
$$ [A'B'C']=\frac{1}{2}\left|48+12+24\right|=\color{red}{42} $$
the answer to life, the universe and everything. As an alternative, you may draw $A'B'C'$ and count the lattice points inside it and on its boundary, then apply Pick's theorem to get the same answer. Or: you may compute the side lengths of $A'B'C'$ through the Pythagorean theorem then apply Heron's formula. Or you may look at the area of $A'B'C'$ as the difference between the area of a rectangle ($108$) and the areas of three right triangles ($36+18+12$):

This is the approach I usually take when teaching the shoelace formula.
