Given P(2, -3) and Q(-5, 1) Midpoint = \((\frac{2 + (-5)}{2}, \frac{-3 + 1}{2})\) = \((\frac{-3}{2}, -1)\) Slope of the line PQ = \(\frac{1 - (-3)}{-5 - 2}\) = \(-\frac{4}{7}\) The slope of the perpendicular line to PQ = \(\frac{-1}{-\frac{4}{7}}\) = \(\frac{7}{4}\) The equation of the perpendicular line: \(y = \frac{7}{4}x + b\) Using a point on the line (in this case, the midpoint) to find the value of b (the intercept). \(-1 = (\frac{7}{4})(\frac{-3}{2}) + b\) \(-1 + \frac{21}{8} = \frac{13}{8} = b\) \(\therefore\) The equation of the perpendicular bisector of the line PQ is \(y = \frac{7}{4}x + \frac{13}{8}\) \(\equiv 8y = 14x + 13 \implies 8y - 14x - 13 = 0\)