JAMB Mathematics 2010 Paper

Concept:The locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining them.Explanation:Find the midpoint of $P(1,3)$ and $Q(3,5)$.Midpoint $= \left(\frac{1+3}{2}, \frac{3+5}{2}\right) = (2,4)$.Find the gradient of $PQ$.$m_{PQ} = \frac{5-3}{3-1} = \frac{2}{2} = 1$.The perpendicular bisector has gradient $m = -1$, because the product of perpendicular gradients is $-1$.Use the point-slope form through $(2,4)$ with gradient $-1$.$y - 4 = -1(x - 2)$$y - 4 = -x + 2$$y = -x + 6$Answer:$y = -x + 6$, which is option A.