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
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
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