Aim: How do we find the equation for a compound locus? Do Now ...

Aim: How do we find the equation for acompound locus?Do Now: Write the equation of the locus thatsatisfies the given conditions:(A) abscissas are ­ 5(B) ordinates are ­4(C) ordinate is 3 less than twice the abscissa(D) abscissa and the ordinate have a sum of ­3(E) 3 units away from y = ­2(F) 5 units away from x = 0The locus "n" units away from the linex = b is made up of two lines whoseequations arex = b + n , x = b ­ n

Locus equidistant from two fixed pointsWrite the equation of thelocus equidistant from(a) (­4,2) , (6,2)(b) (5, ­4) , (5, ­10)(c) (0,1), (4,3)87654321­8 ­7 ­6 ­5 ­4 ­3 ­2 ­1­10 1 2 3 4 5 6 7 8­2­3­4­5­6­7­8yxTo find the equation of the perpendicular bisector(1) graph the given points2) Find the slope of the original line segment3) negative reciprocal of answer in (1)

locus a fixed distance from a point1.Write the equation of the locus with thefollowing conditions:(A) 6 units away from the origin(B) √15 units away from (6,­1)(C) 1/2 units away from (5, ­ 4)2.Describe the locus of points for the givenequation:(a) x 2 + y 2 = 17(b) 2x 2 + 2y 2 = 503.Write an equation for the locus that is5 units away from the origin.Tell whether the given point is on the locus(A) (­4,3)(B) (1, ­4)p.603 / 7Write the equation of the locus basedupon the given conditions7. Points in which the abscissa is 3 lessthan the ordinate11. Points eight units away from theorigin13. Points equidistant from 2 lines x =7, x = 1314. Points 3 units from a fixed point(2,6)

P.601 / 28(a) Draw the locus of points equidistant from thepoints (4,1) and (4,5) and write its equation(b) Draw the locus of points equidistant from thepoints (3,2) , and (9,2)(c) Find the number of points that satisfy bothconditions stated in a, b, and give the coordinates ofthese points.y108642x­10 ­8 ­6 ­4 ­2 0 2 4 6 8 10­2­4­6­8­10