JAMB Mathematics 2019 Paper

Concept:The locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining them.Explanation:Since $A(x, y)$ is equidistant from $B(0, 2)$ and $C(2, 1)$, we set $AB = AC$.Squaring both sides gives $(AB)^2 = (AC)^2$.Using the distance formula:$$(x - 0)^2 + (y - 2)^2 = (x - 2)^2 + (y - 1)^2$$Expanding both sides:$$x^2 + y^2 - 4y + 4 = x^2 - 4x + 4 + y^2 - 2y + 1$$Cancel $x^2$, $y^2$, and simplify:$$-4y + 4 = -4x - 2y + 5$$Rearrange to collect like terms:$$4x - 2y = 1$$Answer:The equation of the locus is $4x - 2y = 1$, which is option C.