2013-I-14

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

Therefore, the equation of the straight line perpendicular to $L$ and passing through $R$ is

$\begin{array}{rcl}
y-17 & = & \dfrac{3}{4} ( x-6) \\
4y – 68 & = & 3x-18 \\
3x-4y + 50 & = & 0
\end{array}$

For the coordinates of $P$,

$\left\{ \begin{array}{ll}
4x+3y+50 = 0 & \ldots \unicode{x2460} \\
3x-4y+50 = 0 & \ldots \unicode{x2461}
\end{array}\right.$

$\unicode{x2460}\times 4 +\unicode{x2461}\times 3$, we have

$\begin{array}{rcl}
25x + 350 & = & 0 \\
x & = & -14
\end{array}$

Sub. $x=-14$ into \mycirc{1}, we have

$\begin{array}{rcl}
4(-14)+3y + 50 & = & 0 \\
3y & = & 6 \\
y & = & 2
\end{array}$

Therefore, the coordinates of $P$ is $(-14,2)$. Hence, the distance between $P$ and $R$

$\begin{array}{cl}
= & \sqrt{(-14-6)^2 + ( 2-17)^2} \\
= & 25
\end{array}$

1. $P$, $Q$ and $R$ are collinear.
2. Note that the radius of $C$

   $\begin{array}{cl}
   = & \sqrt{(6)^2+(17)^2 – 225} \\
   = & 10
   \end{array}$

   Therefore, $PQ=15$ and $QR=10$.

   Since $\Delta OPQ$ and $\Delta OQR$ have the same height, then the area of $\Delta OPQ$ : the area of $\Delta OQR$

   $\begin{array}{cl}
   = & PQ : QR \\
   = & 15 : 10 \\
   = & 3 : 2
   \end{array}$