JAMB 2013 · UTME · Q32✱✱

Find the equation of the perpendicular bisector of the line joining $P(2, 3)$ to $Q(-5, 1)$ .

Worked solution (try it first)

Midpoint of $PQ$ : $\left(\frac{2 + (-5)}{2}, \frac{3 + 1}{2}\right) = \left(-\frac32, 2\right)$ .
Gradient of $PQ$ : $\frac{1 - 3}{-5 - 2} = \frac{-2}{-7}$
$= \frac27$ .
The perpendicular gradient is its negative reciprocal, $-\frac72$ .
Line through the midpoint: $y - 2 = -\frac72\left(x + \frac32\right)$ .
Multiply by 4: $4y - 8 = -14x - 21$ .
Bring everything to the left: $4y + 14x + 13 = 0$ , option A.