JAMB Mathematics 2013 Paper

Given P(2, -3) and Q(-5, 1)

Midpoint = $\left( \frac{2 + (-5)}{2}, \frac{-3 + 1}{2} \right)$

= $\left( \frac{-3}{2}, -1 \right)$

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).

$-1 = \left( \frac{7}{4} \right) \left( \frac{-3}{2} \right) + b$

$-1 + \frac{21}{8} = \frac{13}{8} = b$

$\therefore$ The equation of the perpendicular bisector of the line PQ is $y = \frac{7}{4}x + \frac{13}{8}$

$\equiv 8y = 14x + 13 \implies 8y - 14x - 13 = 0$