JAMB 2010 · UTME · Q34

What is the locus of a point that is equidistant from the points $P(1, 3)$ and $Q(3, 5)$ ?

Worked solution (try it first)

Points equidistant from $P$ and $Q$ lie on the perpendicular bisector of $PQ$ .
The mid-point of $PQ$ is $\left(\frac{1 + 3}{2}, \frac{3 + 5}{2}\right) = (2, 4)$ .
The gradient of $PQ$ is $\frac{5 - 3}{3 - 1} = 1$ .
Perpendicular lines have gradients that multiply to $-1$ , so the bisector has gradient $-1$ .
Through $(2, 4)$ with gradient $-1$ : $y - 4 = -(x - 2)$ , so $y = -x + 6$ , option A.