12. $A B C D$ is a square on the rectangular coordinate plane, and $(31,27),(42,43)$, $(60,27)$ and $(46,16)$ are points on its sides $A B, B C, C D$ and $D A$ respectively. Find the area of $A B C D$.
(2 marks)
$A B C D$ 是直角座標平面上的一個正方形, 而 $(31,27)$ 、 $(42,43)$ 、 $(60,27)$ 和 $(46,16)$ 分別是邊 $A B 、 B C 、 C D$ 和 $D A$ 上的點。求 $A B C D$ 的面積。
(2 分)
12. $\frac{3364}{5}$
12. Let the four given points be $P, Q, R, S$ in order.
Observe that $P R$ is horizontal with length 29.
Hence if we draw a vertical line segment from $Q$
to meet the square at $T$, the length of $Q T$ will be 29
as well (think of rotation everything by $90^{\circ}$ ). Thus
$T$ has coordinates $(42,14)$. With coordinates of $S$
and $T$, we know that $A D$ has slope 0.5 while $B A$
and $C D$ have slope -2 . We can then find that the
equations of $A D, A B$ and $C D$ are $x-2 y-14=0$,
$y=-2 x+89$ and $y=-2 x+147$ respectively.

Solving the first two equations gives the coordinates of $A$ to be $(38.4,12.2)$, while solving the first and third equations gives the coordinates of $D$ to be $(61.8,23.8)$. It follows that the area of $A B C D$ is $A D^{2}=(61.6-38.4)^{2}+(23.8-12.2)^{2}=672.8$.