JAMB Mathematics 2017 Paper

Concept:The point $P(x,y)$ is equidistant from $V(1,1)$ and $W(3,5)$, so its locus is the perpendicular bisector of line segment $VW$.Explanation:Write the distance condition $PV = PW$ using the distance formula.$$\sqrt{(x-1)^2 + (y-1)^2} = \sqrt{(x-3)^2 + (y-5)^2}$$Square both sides to remove the square roots.$$(x-1)^2 + (y-1)^2 = (x-3)^2 + (y-5)^2$$Expand each expression.$$x^2 - 2x + 1 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25$$Simplify both sides.$$x^2 + y^2 - 2x - 2y + 2 = x^2 + y^2 - 6x - 10y + 34$$Cancel $x^2$ and $y^2$, then collect like terms.$$-2x + 6x - 2y + 10y = 34 - 2$$$$4x + 8y = 32$$Divide through by $4$.$$x + 2y = 8$$Answer:The equation of the locus is $x + 2y = 8$, which is option D.