2011SP-I-13

The equation of $L_2$ is

$\begin{array}{rcl}
y-9 & = & -\dfrac{3}{4}(x-4) \\
4y-36 & = & -3x+12 \\
3x+4y-48 & = & 0
\end{array}$

1. Since $\Gamma$ is the perpendicular bisector of $AB$, $\Gamma$ is parallel to $L_2$.
2. Note that the slope of $\Gamma$ is $-\dfrac{3}{4}$.$\left\{ \begin{array}{ll}
   4x-3y+12=0 & \ldots\unicode{x2460} \\
   3x+4y-48=0 & \ldots \unicode{x2461}
   \end{array} \right. $

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

   $\begin{array}{rcl}
   25x-96 & = & 0 \\
   x & = & \dfrac{96}{25}
   \end{array}$

   Sub. $x=\dfrac{96}{25}$ into $\unicode{x2461}$, we have

   $\begin{array}{rcl}
   3(\dfrac{96}{25}) +4y-48 & = & 0 \\
   y & = & \dfrac{228}{25}
   \end{array}$

   $\therefore A=(\dfrac{96}{25},\dfrac{228}{25})$.

   Sub. $x=0$ into the equation of $L_1$, we have $y=4$. Therefore $B=(0,4)$.

   $\therefore$ the mid-point of $AB$

   $\begin{array}{cl}
   = & (\dfrac{0+\frac{96}{25}}{2}, \dfrac{4+\frac{228}{25}}{2}) \\
   = & (\dfrac{48}{25}, \dfrac{164}{25})
   \end{array}$

   $\therefore$ the equation of $\Gamma$ is

   $\begin{array}{rcl}
   y-\dfrac{164}{25} & = & -\dfrac{3}{4} (x-\dfrac{48}{25}) \\
   100y-656 & = & -75x+144\\
   3x+4y-32 & = & 0
   \end{array}$