AMSCO'S Geometry. New York - Rye High School

620 Locus and Constructionmidpoint of ABslope of5 A 2 1 82 , 5 1 (21)2 B5 (5, 2)AB 5 5 2 (21)8 2 2yB(8, 5)5 6 65 1Therefore, the slope of a line perpendicularto AB is 1. The perpendicular bisector ofAB is the line through (5, 2) with slope 1.The equation of this line is:1Oy = x 71A(2, 1)(5, 2)xy 2 2x 2 5 521y 2 2 52x 1 5y 52x 1 7EXAMPLE 1Describe and write an equation for the locus of points equidistant fromA(2, 5) and B(6, 1).Solution (1) Find the midpoint, M, of AB :M A 22 2 1 6 , 5 1 2 1 B (2, 3)(2) Find the slope of AB :slope ofAB 5 5 2 122 2 6 5 4 28 521 2(3) The slope of a line perpendicular toAB is 2.(4) Write an equation of the line through(2, 3) with slope 2.y 2 3x 2 2 5 2y 2 3 5 2x 2 4y 5 2x 2 1A(2, 5)y = 2x 1y1O 1M(2, 3)B(6, 1)xAnswerThe locus of points equidistant from A(2, 5) and B(6, 1) is the perpendicularbisector of AB. The equation of the locus is y 2x 1.