A perpendicular bisector to is drawn through point C and extended to point D. Arrange the pairs of points in order of the y-intercepts of their corresponding perpendicular bisectors, starting with the smallest and ending with the largest.
Answers
Sample perpendicular bisector diagram:
Solution:
Pair 1: slope = (9 - 5)/(8+4) = 1/3: midpoint = ((-4+8)/2, (5+9)/2) = (2, 7): perpendicular bisector passes through point (2, 7) with slope = -1/(1/3) = -3 giving the equation (y - 7)/(x - 2) = -3 or y - 7 = -3(x - 2) or y = -3x + 13 and y-intercept at y = 13.:
Pair 2: slope = (6 - 4)/(-8-2) = -1/5:midpoint = ((2-8)/2, (4+6)/2) = (-3, 5):perpendicular bisector passes through point (-3, 5) with slope = -1/(-1/5) = 5 giving the equation (y -5)/(x + 3) = 5 or y - 5 = 5(x + 3) or y = 5x + 20 and y-intercept at y = 20.
Pair 3: slope = (2 - 4)/(7 - 5) = -1:midpoint = ((5+7)/2, (4+2)/2) = (6, 3):perpendicular bisector passes through point (6, 3) with slope = -1/(-1) = 1 giving the equation (y - 3)/(x - 6) = 1 or y - 3 = (x - 6) or y = x - 3 and y-intercept at y = -3.:
Pair 4: slope = (3 - 9)/(-4 - 2) = 1:midpoint = ((2-4)/2, (9+3)/2) = (-1, 6):perpendicular bisector passes through point (-1, 6) with slope = -1(1) = -1 giving the equation (y - 6)/(x + 1) = -1 or y - 6 = -1(x + 1) or y = -x + 5 and y-intercept at y = 5.
Pair 5: slope = (-12 + 2)/(9 - 3) = -5/3:midpoint = ((3+9)/2, (-2-12)/2) = (6, -7):perpendicular bisector passes through point (6, -7) with slope = -1(-5/3) = 3/5 giving the equation (y + 7)/(x - 6) = 3/5 or 5(y + 7) = 3(x - 6) or 5y = 3x - 53 and y-intercept at y = -10.6.
Pair 6: slope = (12 - 10)/(8 - 4) = ½:midpoint = ((4+8)/2, (10+12)/2) = (6, 11):perpendicular bisector passes through point (6, 11) with slope = -1(1/2) = -2 giving the equation (y - 11)/(x - 6) = -2 or y - 11 = -2(x - 6) or y = -2x + 23 and y-intercept at y = 23.
Arrangement in order of y-intercepts from smallest to largest: a(3, -2) and b(9, -12): a(5, 4) and b(7, 2): a(2, 9) and b(-4, 3): a(-4, 5) and b(8, 9): a(2, 4) and b(-8, 6): a(4, 10) and b(8, 12).