Medians and Altitudes of Triangles

1. SECTION 5-2
Medians and Altitudes of Triangles
Thursday, March 1, 2012

2. ESSENTIAL QUESTIONS
How do you identify and use medians in triangles?
How do you identify and use altitudes in triangles?
Thursday, March 1, 2012

3. VOCABULARY
1. Median:
2. Centroid:
3. Altitude:
4. Orthocenter:
Thursday, March 1, 2012

4. VOCABULARY
1. Median: A segment in a triangle that connects a vertex to
the midpoint of the opposite side
2. Centroid:
3. Altitude:
4. Orthocenter:
Thursday, March 1, 2012

5. VOCABULARY
1. Median: A segment in a triangle that connects a vertex to
the midpoint of the opposite side
2. Centroid: The point of concurrency where the medians
intersect in a triangle (always inside)
3. Altitude:
4. Orthocenter:
Thursday, March 1, 2012

6. VOCABULARY
1. Median: A segment in a triangle that connects a vertex to
the midpoint of the opposite side
2. Centroid: The point of concurrency where the medians
intersect in a triangle (always inside)
3. Altitude: A segment in a triangle that connects a vertex to
the opposite side so that the side and altitude are
perpendicular
4. Orthocenter:
Thursday, March 1, 2012

7. VOCABULARY
1. Median: A segment in a triangle that connects a vertex to
the midpoint of the opposite side
2. Centroid: The point of concurrency where the medians
intersect in a triangle (always inside)
3. Altitude: A segment in a triangle that connects a vertex to
the opposite side so that the side and altitude are
perpendicular
4. Orthocenter: The point of concurrency where the altitudes
of a triangle intersect
Thursday, March 1, 2012

8. 5.7 - CENTROID THEOREM
The centroid of a triangle is two-thirds the distance from each vertex
to the midpoint of the opposite side
Thursday, March 1, 2012

9. 5.7 - CENTROID THEOREM
The centroid of a triangle is two-thirds the distance from each vertex
to the midpoint of the opposite side
If G is the centroid of ∆ABC, then
2 2 2
AG = AF , BG = BD, and CG = CE
3 3 3
Thursday, March 1, 2012

10. Special Segments and Points in Triangles
Point of Special
Name Example Example
Concurrency Property
Perpendicular Circumcenter is
Circumcenter equidistant from
Bisector each vertex
Angle In center is
Incenter equidistant from
Bisector each side
Centroid is two-
thirds the distance
Median Centroid from vertex to
opposite midpoint
Altitudes are
Altitude Orthocenter concurrent at
orthocenter
Thursday, March 1, 2012

11. EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
Thursday, March 1, 2012

12. EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
2
YP = YV
3
Thursday, March 1, 2012

13. EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
2
YP = YV
3
2
YP = (12)
3
Thursday, March 1, 2012

14. EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
2
YP = YV
3
2
YP = (12)
3
YP = 8
Thursday, March 1, 2012

15. EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
2
YP = YV PV = YV − YP
3
2
YP = (12)
3
YP = 8
Thursday, March 1, 2012

16. EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
2
YP = YV PV = YV − YP
3
2
YP = (12) PV = 12 − 8
3
YP = 8
Thursday, March 1, 2012

17. EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
2
YP = YV PV = YV − YP
3
2
YP = (12) PV = 12 − 8
3
YP = 8 PV = 4
Thursday, March 1, 2012

18. EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
Thursday, March 1, 2012

19. EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
2
CG = CE
3
Thursday, March 1, 2012

20. EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
2
CG = CE
3
2
4 = CE
3
Thursday, March 1, 2012

21. EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
2
CG = CE
3
2
4 = CE
3
CE = 6
Thursday, March 1, 2012

22. EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
2
CG = CE GE = CE − CG
3
2
4 = CE
3
CE = 6
Thursday, March 1, 2012

23. EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
2
CG = CE GE = CE − CG
3
2
4 = CE GE = 6 − 4
3
CE = 6
Thursday, March 1, 2012

24. EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
2
CG = CE GE = CE − CG
3
2
4 = CE GE = 6 − 4
3
CE = 6 GE = 2
Thursday, March 1, 2012

25. EXAMPLE 3
An artist is designing a sculpture that balances a triangle on
tip of a pole. In the artistʼs design on the coordinate plane,
the vertices are located at (1, 4), (3, 0), and (3, 8). What are
the coordinates of the point where the artist should place
the pole under the triangle so that is will balance?
Thursday, March 1, 2012

26. EXAMPLE 3
(1, 4), (3, 0), (3, 8)
Thursday, March 1, 2012

27. EXAMPLE 3
y
(1, 4), (3, 0), (3, 8)
x
Thursday, March 1, 2012

28. EXAMPLE 3
y
(1, 4), (3, 0), (3, 8)
x
Thursday, March 1, 2012

29. EXAMPLE 3
y
(1, 4), (3, 0), (3, 8)
x
Thursday, March 1, 2012

30. EXAMPLE 3
y
(1, 4), (3, 0), (3, 8)
We need to find the centroid, so
we start by finding the midpoint
of our vertical side.
x
Thursday, March 1, 2012

31. EXAMPLE 3
y
(1, 4), (3, 0), (3, 8)
We need to find the centroid, so
we start by finding the midpoint
of our vertical side.
x
⎛ 3 + 3 0 + 8⎞
M =⎜ , ⎟
⎝ 2 2 ⎠
Thursday, March 1, 2012

32. EXAMPLE 3
y
(1, 4), (3, 0), (3, 8)
We need to find the centroid, so
we start by finding the midpoint
of our vertical side.
x
⎛ 3 + 3 0 + 8⎞ ⎛ 6 8⎞
M =⎜ , ⎟ = ⎜ 2 , 2⎟
⎝ 2 2 ⎠ ⎝ ⎠
Thursday, March 1, 2012

33. EXAMPLE 3
y
(1, 4), (3, 0), (3, 8)
We need to find the centroid, so
we start by finding the midpoint
of our vertical side.
x
⎛ 3 + 3 0 + 8⎞ ⎛ 6 8⎞
M =⎜ , ⎟ = ⎜ 2 , 2⎟
⎝ 2 2 ⎠ ⎝ ⎠
( )
= 3,4
Thursday, March 1, 2012

34. EXAMPLE 3
y
(1, 4), (3, 0), (3, 8)
We need to find the centroid, so
we start by finding the midpoint
of our vertical side.
x
⎛ 3 + 3 0 + 8⎞ ⎛ 6 8⎞
M =⎜ , ⎟ = ⎜ 2 , 2⎟
⎝ 2 2 ⎠ ⎝ ⎠
( )
= 3,4
Thursday, March 1, 2012

35. EXAMPLE 3
y
(1, 4), (3, 0), (3, 8)
We need to find the centroid, so
we start by finding the midpoint
of our vertical side.
x
⎛ 3 + 3 0 + 8⎞ ⎛ 6 8⎞
M =⎜ , ⎟ = ⎜ 2 , 2⎟
⎝ 2 2 ⎠ ⎝ ⎠
( )
= 3,4
Thursday, March 1, 2012

36. EXAMPLE 3
y
x
Thursday, March 1, 2012

37. EXAMPLE 3
y
Next, we need the distance from
the opposite vertex to this
midpoint.
x
Thursday, March 1, 2012

38. EXAMPLE 3
y
Next, we need the distance from
the opposite vertex to this
midpoint.
d = (1 − 3)2 + (4 − 4)2
x
Thursday, March 1, 2012

39. EXAMPLE 3
y
Next, we need the distance from
the opposite vertex to this
midpoint.
d = (1 − 3)2 + (4 − 4)2
x
= (−2) + 0
2 2
Thursday, March 1, 2012

40. EXAMPLE 3
y
Next, we need the distance from
the opposite vertex to this
midpoint.
d = (1 − 3)2 + (4 − 4)2
x
= (−2) + 0
2 2
= 4
Thursday, March 1, 2012

41. EXAMPLE 3
y
Next, we need the distance from
the opposite vertex to this
midpoint.
d = (1 − 3)2 + (4 − 4)2
x
= (−2) + 0
2 2
= 4 =2
Thursday, March 1, 2012

42. EXAMPLE 3
y
x
Thursday, March 1, 2012

43. EXAMPLE 3
y
The centroid P is 2/3 of this
distance from the vertex.
x
Thursday, March 1, 2012

44. EXAMPLE 3
y
The centroid P is 2/3 of this
distance from the vertex.
⎛ 2 ⎞
P = ⎜1 + (2),4 ⎟
⎝ 3 ⎠
x
Thursday, March 1, 2012

45. EXAMPLE 3
y
The centroid P is 2/3 of this
distance from the vertex.
⎛ 2 ⎞
P = ⎜1 + (2),4 ⎟
⎝ 3 ⎠
x ⎛ 4 ⎞
P = ⎜1 + ,4 ⎟
⎝ 3 ⎠
Thursday, March 1, 2012

46. EXAMPLE 3
y
The centroid P is 2/3 of this
distance from the vertex.
⎛ 2 ⎞
P = ⎜1 + (2),4 ⎟
⎝ 3 ⎠
x ⎛ 4 ⎞
P = ⎜1 + ,4 ⎟
⎝ 3 ⎠
⎛7 ⎞
P = ⎜ ,4 ⎟
⎝3 ⎠
Thursday, March 1, 2012

47. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
Thursday, March 1, 2012

48. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
x
Thursday, March 1, 2012

49. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
H
J
x
I
Thursday, March 1, 2012

50. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
H
J
x
I
Thursday, March 1, 2012

51. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
H
J
x
I
Thursday, March 1, 2012

52. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
To find the orthocenter, find the
intersection of two altitudes.
H
J
x
I
Thursday, March 1, 2012

53. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
To find the orthocenter, find the
intersection of two altitudes.
H
J Let’s find the equations for the
altitudes coming from I and H.
x
I
Thursday, March 1, 2012

54. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
To find the orthocenter, find the
intersection of two altitudes.
H
J Let’s find the equations for the
altitudes coming from I and H.
x
I ( )
m JI =
1+ 3
−5 + 3
Thursday, March 1, 2012

55. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
To find the orthocenter, find the
intersection of two altitudes.
H
J Let’s find the equations for the
altitudes coming from I and H.
x
I ( )
m JI =
1+ 3
=
4
−5 + 3 −2
Thursday, March 1, 2012

56. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
To find the orthocenter, find the
intersection of two altitudes.
H
J Let’s find the equations for the
altitudes coming from I and H.
x
I ( )
m JI =
1+ 3
=
4
−5 + 3 −2
= −2
Thursday, March 1, 2012

57. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
To find the orthocenter, find the
intersection of two altitudes.
H
J Let’s find the equations for the
altitudes coming from I and H.
x
I ( )
m JI =
1+ 3
=
4
−5 + 3 −2
= −2 ⊥ m =
1
2
Thursday, March 1, 2012

58. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
To find the orthocenter, find the
intersection of two altitudes.
H
J Let’s find the equations for the
altitudes coming from I and H.
x
I ( )
m JI =
1+ 3
=
4
−5 + 3 −2
= −2 ⊥ m =
1
2
( )
m HJ =
2 −1
1+ 5
Thursday, March 1, 2012

59. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
To find the orthocenter, find the
intersection of two altitudes.
H
J Let’s find the equations for the
altitudes coming from I and H.
x
I ( )
m JI =
1+ 3
=
−5 + 3 −2
4
= −2 ⊥ m =
1
2
( )
m HJ =
2 −1 1
1+ 5 6
=
Thursday, March 1, 2012

60. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
To find the orthocenter, find the
intersection of two altitudes.
H
J Let’s find the equations for the
altitudes coming from I and H.
x
I ( )
m JI =
1+ 3
=
−5 + 3 −2
4
= −2 ⊥ m =
1
2
( )
m HJ =
2 −1 1
1+ 5 6
= ⊥ m = −6
Thursday, March 1, 2012

61. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
Altitude through H
H
J
x
I
Thursday, March 1, 2012

62. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
Altitude through H
1
⊥ m = , (1,2)
H 2
J
x
I
Thursday, March 1, 2012

63. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
Altitude through H
1
⊥ m = , (1,2)
H 2
J 1
y − 2 = (x −1)
2
x
I
Thursday, March 1, 2012

64. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
Altitude through H
1
⊥ m = , (1,2)
H 2
J 1
y − 2 = (x −1)
2
x 1 1
I y−2= x −
2 2
Thursday, March 1, 2012

65. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
Altitude through H
1
⊥ m = , (1,2)
H 2
J 1
y − 2 = (x −1)
2
x 1 1
I y−2= x −
2 2
1 3
y= x+
2 2
Thursday, March 1, 2012

66. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
Altitude through I
H
J
x
I
Thursday, March 1, 2012

67. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
Altitude through I
⊥ m = −6, (−3,−3)
H
J
x
I
Thursday, March 1, 2012

68. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
Altitude through I
⊥ m = −6, (−3,−3)
H y + 3 = −6(x + 3)
J
x
I
Thursday, March 1, 2012

69. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
Altitude through I
⊥ m = −6, (−3,−3)
H y + 3 = −6(x + 3)
J
x y + 3 = −6x −18
I
Thursday, March 1, 2012

70. EXAMPLE 4
The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
y
Altitude through I
⊥ m = −6, (−3,−3)
H y + 3 = −6(x + 3)
J
x y + 3 = −6x −18
I y = −6x − 21
Thursday, March 1, 2012

71. EXAMPLE 4
Thursday, March 1, 2012

72. EXAMPLE 4
1 3
−6x − 21 = x +
2 2
Thursday, March 1, 2012

73. EXAMPLE 4
1 3
−6x − 21 = x +
2 2
+6x +6x
Thursday, March 1, 2012

74. EXAMPLE 4
1 3
−6x − 21 = x +
2 2
+6x +6x
13 3
−21 = x +
2 2
Thursday, March 1, 2012

75. EXAMPLE 4
1 3
−6x − 21 = x +
2 2
+6x +6x
13 3
−21 = x +
2 2
3 3
− −
2 2
Thursday, March 1, 2012

76. EXAMPLE 4
1 3
−6x − 21 = x +
2 2
+6x +6x
13 3
−21 = x +
2 2
3 3
− −
2 2
45 13
− = x
2 2
Thursday, March 1, 2012

77. EXAMPLE 4
1 3 2 ⎛ 45 ⎞ ⎛ 13 ⎞ 2
−6x − 21 = x + ⎜ − 2 ⎟ = ⎜ 2 x ⎟ 13
2 2 13 ⎝ ⎠ ⎝ ⎠
+6x +6x
13 3
−21 = x +
2 2
3 3
− −
2 2
45 13
− = x
2 2
Thursday, March 1, 2012

78. EXAMPLE 4
1 3 2 ⎛ 45 ⎞ ⎛ 13 ⎞ 2
−6x − 21 = x + ⎜ − 2 ⎟ = ⎜ 2 x ⎟ 13
2 2 13 ⎝ ⎠ ⎝ ⎠
+6x +6x
45
13 3 x=−
−21 = x + 13
2 2
3 3
− −
2 2
45 13
− = x
2 2
Thursday, March 1, 2012

79. EXAMPLE 4
1 3 2 ⎛ 45 ⎞ ⎛ 13 ⎞ 2
−6x − 21 = x + ⎜ − 2 ⎟ = ⎜ 2 x ⎟ 13
2 2 13 ⎝ ⎠ ⎝ ⎠
+6x +6x
45
13 3 x=−
−21 = x + 13
2 2
⎛ 45 ⎞ 1 ⎛ 45 ⎞ 3
3 3 −6 ⎜ − ⎟ − 21 = ⎜ − ⎟ +
− − ⎝ 13 ⎠ 2 ⎝ 13 ⎠ 2
2 2
45 13
− = x
2 2
Thursday, March 1, 2012

80. EXAMPLE 4
1 3 2 ⎛ 45 ⎞ ⎛ 13 ⎞ 2
−6x − 21 = x + ⎜ − 2 ⎟ = ⎜ 2 x ⎟ 13
2 2 13 ⎝ ⎠ ⎝ ⎠
+6x +6x
45
13 3 x=−
−21 = x + 13
2 2
⎛ 45 ⎞ 1 ⎛ 45 ⎞ 3
3 3 −6 ⎜ − ⎟ − 21 = ⎜ − ⎟ +
− − ⎝ 13 ⎠ 2 ⎝ 13 ⎠ 2
2 2
45 13 3 3
− = x − =−
2 2 13 13
Thursday, March 1, 2012

81. EXAMPLE 4
1 3 2 ⎛ 45 ⎞ ⎛ 13 ⎞ 2
−6x − 21 = x + ⎜ − 2 ⎟ = ⎜ 2 x ⎟ 13
2 2 13 ⎝ ⎠ ⎝ ⎠
+6x +6x
45
13 3 x=−
−21 = x + 13
2 2
⎛ 45 ⎞ 1 ⎛ 45 ⎞ 3
3 3 −6 ⎜ − ⎟ − 21 = ⎜ − ⎟ +
− − ⎝ 13 ⎠ 2 ⎝ 13 ⎠ 2
2 2
45 13 3 3
− = x − =−
2 2 13 13 ⎛ 45 3 ⎞
⎜ − 13 ,− 13 ⎟
⎝ ⎠
Thursday, March 1, 2012

82. CHECK YOUR
UNDERSTANDING
p. 337 #1-4
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83. PROBLEM SET
Thursday, March 1, 2012

84. PROBLEM SET
p. 338 #5-31 odd, 49, 53
"Education's purpose is to replace an empty mind with an open
one." – Malcolm Forbes
Thursday, March 1, 2012