2007-I-13

$\begin{array}{rcl}
y-3 & = & \dfrac{-4}{3} (x-10) \\
3(y-3) & = & -4(x-10) \\
3y-9 & = & -4x +40 \\
4x + 3y – 49 & = & 0
\end{array}$

$\begin{array}{rcl}
4(4) + 3(h) -49 & = & 0 \\
3h – 33 & = & 0 \\
3h & = & 33 \\
h & = & 11
\end{array}$

1. 留意 $AB=AC$。$\Delta ABC$ 為一等腰三角形。所以通過 $A(4,h)$ 的鉛垂線 $x=4$ 為 $\Delta ABC$ 的對稱軸。由此，可得

   $\begin{array}{rcl}
   \dfrac{10+ k}{2} & = & 4 \\
   10 + k & = & 8 \\
   k & = & -2
   \end{array}$

2. 以 $BC$ 為底的話，$\Delta ABC$ 的高為

   $\begin{array}{cl}
   = & 11 – 3 \\
   = & 8
   \end{array}$

   及 $BC$

   $\begin{array}{cl}
   = & 10 – (-2) \\
   = & 12
   \end{array}$

   所以，$\Delta ABC$ 的面積

   $\begin{array}{cl}
   = & \dfrac{1}{2} \times 12 \times 8 \\
   = & 48
   \end{array}$

   以 $AC$ 為底的話，$BD$ 則是 $\Delta ABC$ 的高。$AB$

   $\begin{array}{cl}
   = & \sqrt{(4-(-2))^2 + (11-3)^2} \\
   = & 10
   \end{array}$

   由此，可得

   $\begin{array}{rcl}
   \dfrac{1}{2} \times 10 \times BD & = & 48 \\
   BD & = & \dfrac{48}{5}
   \end{array}$