Question

homework help example 1: exs. 16 - 10 example 2: exs. 20 - 23 example 3: exs. 26 - 29 example 4: exs. 30 - 35 finding segment lengths m is the mid - point of the segment. find the segment lengths. 16. find km and ml. 17. find dm and me. 18. find ym and mz. 19. find am and mb. finding segment lengths line l bisects the segment. find the segment lengths. 20. find cb and ab. 21. find mp and mn. 22. find fe and de. 23. find ut and st. biking the minuteman bikeway is a 10.5 mile bike path that runs from arlington to bedford, massachusetts. 24. caitlin and laurie begin at opposite ends of the minuteman bikeway and meet at the halfway point on the path. how far does each rider bike? 25. caitlin starts on the path 4.3 miles from the arlington end. laurie starts on the path 3 miles from the bedford end. how far will each rider bike before reaching the halfway point on the path? 26. using algebra find the value of the variable. 27. 28. 29. midpoint formula find the coordinates of the mid - point of pq. 30. p(0, 0), q(4, 6) 31. p(3, 8), q(7, 6) 32. p(-5, 6), q(9, 7) 33. p(-12, 0), q(6, 1) 34. p(-4, 4), q(4, 0) 35. p(3, 2), q(-7, -4) 2.1 segment bisectors 57 student help classzone.com homework help extra help with problem solving in exs. 30 - 35 is at classzone.com

Explanation:

Step1: Recall mid - point property

If \(M\) is the mid - point of a line segment \(AB\), then \(AM = MB=\frac{AB}{2}\). If a line \(l\) bisects a line segment \(AB\) at point \(C\), then \(AC = CB=\frac{AB}{2}\).

Step2: Solve problem 16

Since \(M\) is the mid - point of \(KL\) and \(KL = 30\), then \(KM=ML=\frac{30}{2}=15\).

Step3: Solve problem 17

Since \(M\) is the mid - point of \(DE\) and \(DE = 82\), then \(DM = ME=\frac{82}{2}=41\).

Step4: Solve problem 18

Since \(M\) is the mid - point of \(YZ\) and \(YZ = 17\), then \(YM = MZ=\frac{17}{2}=8.5\).

Step5: Solve problem 19

Since \(M\) is the mid - point of \(AB\) and \(AB = 2.7\), then \(AM = MB=\frac{2.7}{2}=1.35\).

Step6: Solve problem 20

Since line \(l\) bisects \(AB\) at \(C\) and \(AC = 36\), then \(CB=AC = 36\) and \(AB=2\times AC=72\).

Step7: Solve problem 21

Since line \(l\) bisects \(MN\) at \(P\) and \(PN = 15\), then \(MP=PN = 15\) and \(MN = 2\times PN=30\).

Step8: Solve problem 22

Since line \(l\) bisects \(DE\) at \(F\) and \(DF = 29.5\), then \(FE=DF = 29.5\) and \(DE=2\times DF = 59\).

Step9: Solve problem 23

Since line \(l\) bisects \(ST\) at \(U\) and \(SU = 3.6\), then \(UT=SU = 3.6\) and \(ST=2\times SU=7.2\).

Step10: Solve problem 24

The length of the Minuteman Bikeway is \(10.5\) miles. Since they meet at the halfway point, each rider bikes \(\frac{10.5}{2}=5.25\) miles.

Step11: Solve problem 25

The halfway point of the \(10.5\) - mile path is at \(10.5\div2 = 5.25\) miles from either end.
Caitlin starts \(4.3\) miles from the Arlington end. So she bikes \(5.25−4.3 = 0.95\) miles.
Laurie starts \(3\) miles from the Bedford end. So she bikes \(5.25 - 3=2.25\) miles.

Step12: Solve problem 26

Since \(M\) is the mid - point of \(AB\), then \(6p=72\), so \(p = 12\).

Step13: Solve problem 27

Since \(M\) is the mid - point of \(AB\), then \(19=q + 7\), so \(q=19 - 7=12\).

Step14: Solve problem 28

Since \(M\) is the mid - point of \(AB\), then \(r−3 = 15\), so \(r=15 + 3=18\).

Step15: Solve problem 29

Since \(M\) is the mid - point of \(AB\), then \(4=2s+6\), \(2s=4 - 6=-2\), so \(s=-1\).

Step16: Solve problem 30

The mid - point formula for two points \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\). For \(P(0,0)\) and \(Q(4,6)\), the mid - point is \((\frac{0 + 4}{2},\frac{0+6}{2})=(2,3)\).

Step17: Solve problem 31

For \(P(3,8)\) and \(Q(7,6)\), the mid - point is \((\frac{3 + 7}{2},\frac{8 + 6}{2})=(5,7)\).

Step18: Solve problem 32

For \(P(-5,6)\) and \(Q(9,7)\), the mid - point is \((\frac{-5 + 9}{2},\frac{6+7}{2})=(2,6.5)\).

Step19: Solve problem 33

For \(P(-12,0)\) and \(Q(6,1)\), the mid - point is \((\frac{-12 + 6}{2},\frac{0 + 1}{2})=(-3,0.5)\).

Step20: Solve problem 34

For \(P(-4,4)\) and \(Q(4,0)\), the mid - point is \((\frac{-4 + 4}{2},\frac{4+0}{2})=(0,2)\).

Step21: Solve problem 35

For \(P(3,2)\) and \(Q(-7,-4)\), the mid - point is \((\frac{3-7}{2},\frac{2-4}{2})=(-2,-1)\).

Answer:

1. \(KM = 15\), \(ML = 15\)
2. \(DM = 41\), \(ME = 41\)
3. \(YM = 8.5\), \(MZ = 8.5\)
4. \(AM = 1.35\), \(MB = 1.35\)
5. \(CB = 36\), \(AB = 72\)
6. \(MP = 15\), \(MN = 30\)
7. \(FE = 29.5\), \(DE = 59\)
8. \(UT = 3.6\), \(ST = 7.2\)
9. Each rider bikes \(5.25\) miles
10. Caitlin bikes \(0.95\) miles, Laurie bikes \(2.25\) miles
11. \(p = 12\)
12. \(q = 12\)
13. \(r = 18\)
14. \(s=-1\)
15. \((2,3)\)
16. \((5,7)\)
17. \((2,6.5)\)
18. \((-3,0.5)\)
19. \((0,2)\)
20. \((-2,-1)\)