\begin{tabular}{l} lie on the perpendicular bisector of \( \overline{Q R} \) with the \\ \( \begin{array}{ll}\text { 9. } Q(-5,4), R(8,-3), S(-2,-5) \\ 5)=1\end{array} \) \\ \hline\end{tabular}

To find out if point \( S(-2,-5) \) lies on the perpendicular bisector of \( \overline{QR} \), we first need to determine the midpoint of \( \overline{QR} \) and the slope of line \( QR \). The coordinates of \( Q \) are \( (-5,4) \) and \( R \) are \( (8,-3) \). The midpoint \( M \) of \( \overline{QR} \) is found using the formula: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{-5 + 8}{2}, \frac{4 - 3}{2} \right) = \left( \frac{3}{2}, \frac{1}{2} \right). \] Next, we calculate the slope of \( QR \): \[ \text{slope of } QR = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 4}{8 - (-5)} = \frac{-7}{13}. \] The slope of the perpendicular bisector is the negative reciprocal, which is \( \frac{13}{7} \). Now, we use the point-slope form of the equation of a line to derive the equation of the perpendicular bisector: \[ y - y_1 = m(x - x_1) \Rightarrow y - \frac{1}{2} = \frac{13}{7}\left(x - \frac{3}{2}\right). \] This simplifies to: \[ y = \frac{13}{7}x - \frac{39}{14} + \frac{7}{14} = \frac{13}{7}x - \frac{32}{14} = \frac{13}{7}x - \frac{16}{7}. \] Finally, we check if point \( S(-2,-5) \) satisfies this equation: Substituting \( x = -2 \): \[ y = \frac{13}{7}(-2) - \frac{16}{7} = \frac{-26 - 16}{7} = \frac{-42}{7} = -6. \] Since the \( y \)-coordinate of \( S \) is \(-5\) and does not equal \(-6\), point \( S(-2,-5) \) does not lie on the perpendicular bisector of \( \overline{QR} \).