2006-I-12

$\begin{array}{cl}
= & \dfrac{8-0}{12-(-4)} \\
= & \dfrac{1}{2}
\end{array}$

Since $CM \perp AB$, then the slope of $CM$

$\begin{array}{cl}
= & -1 \div \text{the slope of $AB$} \\
= & -1 \div \dfrac{1}{2} \\
= & -2
\end{array}$

Hence, the equation of $CM$ is

$\begin{array}{rcl}
y-4 & = & -2(x-4) \\
y-4 & = & -2x + 8 \\
2x + y -12 & = & 0
\end{array}$

Sub. $y=0$ into the equation of $CM$, we have

$\begin{array}{rcl}
2x +(0) -12 & = & 0 \\
x & = & 6
\end{array}$

Therefore, the coordinates of $C$ is $(6,0)$.

1. The equation of $BD$ is

   $\begin{array}{rcl}
   \dfrac{y-8}{x-12} & = & \dfrac{0-8}{2-12} \\
   \dfrac{y-8}{x-12} & = & \dfrac{4}{5} \\
   5y – 40 & = & 4x – 48 \\
   4x – 5y -8 & = & 0
   \end{array}$

2. $\left\{ \begin{array}{ll} 2x+y-12=0 & \ldots \unicode{x2460} \\
   4x-5y-8=0 & \ldots \unicode{x2461}
   \end{array} \right.$

   $\unicode{x2461} – \unicode{x2460} \times 2$, we have

   $\begin{array}{rcl}
   -7y +16 & = & 0 \\
   y & = & \dfrac{16}{7}
   \end{array}$

   Sub. $y=\dfrac{16}{7}$ into $\unicode{x2460}$, we have

   $\begin{array}{rcl}
   2x + \dfrac{16}{7} -12 & = & 0 \\
   x & = & \dfrac{34}{7}
   \end{array}$

   Therefore, the coordinates of $K$ is $(\dfrac{34}{7}, \dfrac{16}{7})$.

   Consider $\Delta AMC$ and $\Delta AKC$. If we treat $AC$ as the bases of $\Delta AMC$ and $\Delta AKC$, the heights of $\Delta AMC$ and $\Delta AKC$ are the $y$-coordinates of $M$ and $K$ respectively.

   Hence, the area of $\Delta AMC :$ the area of $\Delta AKC$
   $\begin{array}{cl}
   = & \dfrac{1}{2} \times AC \times 4 : \dfrac{1}{2} \times AC \times \dfrac{16}{7} \\
   = & 7 : 4
   \end{array}$