JAMB Mathematics 2013 Paper

Concept:The perpendicular bisector passes through the midpoint of $PQ$ and has a slope that is the negative reciprocal of the slope of $PQ$.Explanation:Midpoint of $P(2,-3)$ and $Q(-5,1)$ is $\left(\frac{2+(-5)}{2}, \frac{-3+1}{2}\right) = \left(-\frac{3}{2}, -1\right)$.Slope of $PQ$ is $\frac{1-(-3)}{-5-2} = \frac{4}{-7} = -\frac{4}{7}$.A perpendicular line has slope $\frac{-1}{-\frac{4}{7}} = \frac{7}{4}$.Let the perpendicular bisector be $y = \frac{7}{4}x + b$.Since it passes through the midpoint, substitute $x = -\frac{3}{2}$ and $y = -1$: $-1 = \frac{7}{4}\left(-\frac{3}{2}\right) + b$.This gives $-1 = -\frac{21}{8} + b$, so $b = \frac{13}{8}$.Thus the equation is $y = \frac{7}{4}x + \frac{13}{8}$.Multiply through by $8$: $8y = 14x + 13$.Rearranging gives $8y - 14x - 13 = 0$.Answer:C. $8y - 14x - 13 = 0$