2008-I-12

$\begin{array}{cl}
= & \dfrac{4-0}{-3-0} \\
= & \dfrac{-4}{3}
\end{array}$

The slope of $OC$

$\begin{array}{cl}
= & \dfrac{-3-0}{4-0} \\
= & \dfrac{-3}{4}
\end{array}$

Since the slope of $OB\neq$ the slope of $OC$, then $O$, $B$ and $C$ are not collinear.

$\begin{array}{cl}
= & \dfrac{4-(-3)}{-3-4} \\
= & -1
\end{array}$

Note that $\angle BCD = 90^\circ$. Hence, we have the slope of $CD$

$\begin{array}{cl}
= & -1 \div \text{the slope of }BC \\
= & -1 \div (-1) \\
= & 1
\end{array}$

Therefore, the equation of $CD$ is

$\begin{array}{rcl}
(y-(-3)) & = & 1 ( x-4) \\
y+3 & = & x – 4 \\
x – y -7 & = & 0
\end{array}$

Note that $A$ is translated horizontally to $D$, then the $y$ coordinates of $A$ and $D$ are equal. Substitute $y=3$ into $x-y-7=0$, we have

$\begin{array}{rcl}
x – (3) – 7 & = & 0 \\
x & = & 10
\end{array}$

Therefore, the coordinates of $D$ are $(10,3)$.