Mathematics 2008 JAMB Past Questions - Question 31: The locus of a point equidistant from two points P (6,2) and R (4,2) is a perpendicular bisector of PR passing through.

Correct Answer

B

Explanation

6+4/2, 2+2/2 = 5,2To find the locus of a point that is equidistant from two given points, we can find the perpendicular bisector of the line segment joining the two points.Given points P(6,2) and R(4,2), we can find the midpoint of the line segment PR by taking the average of the x-coordinates and the average of the y-coordinates:Midpoint = ((6+4)/2, (2+2)/2)        = (5, 2)The midpoint of PR is (5, 2).To find the equation of the perpendicular bisector passing through the midpoint, we need to determine the slope of the line PR. The slope of a line passing through two points (x1, y1) and (x2, y2) is given by:Slope = (y2 - y1) / (x2 - x1)For PR, the slope is:Slope = (2 - 2) / (4 - 6)      = 0 / -2      = 0Since the slope of PR is 0, the slope of the perpendicular bisector will be undefined (or infinite).The equation of the perpendicular bisector passing through the midpoint (5, 2) is of the form x = c, where c is a constant.Therefore, the locus of points equidistant from P(6,2) and R(4,2) is the vertical line x = 5.