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. Note that $AB=AC$. $\Delta ABC$ is an isosceles triangle. Therefore, the vertical line $x=4$ passing through $A(4,h)$ is the axis of the symmetry of $\Delta ABC$. Hence, we have

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

2. With base $BC$, the height of $\Delta ABC$

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

   and $BC$

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

   Therefore, the area of $\Delta ABC$

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

   With base $AC$, $BD$ is the height of $\Delta ABC$. $AB$

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

   Hence, we have

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