SECTION 5-2 Medians and Altitudes of Triangles Thursday, March 1, 2012
SECTION 5-2Medians and Altitudes of Triangles
Thursday, March 1, 2012
ESSENTIAL QUESTIONS
How do you identify and use medians in triangles?
How do you identify and use altitudes in triangles?
Thursday, March 1, 2012
VOCABULARY1. Median:
2. Centroid:
3. Altitude:
4. Orthocenter:
Thursday, March 1, 2012
VOCABULARY1. 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
VOCABULARY1. 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
VOCABULARY1. 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
VOCABULARY1. 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
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
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
AG =
23
AF , BG =23
BD, and CG =23
CE
Thursday, March 1, 2012
Special Segments and Points in TrianglesSpecial Segments and Points in TrianglesSpecial Segments and Points in TrianglesSpecial Segments and Points in TrianglesSpecial Segments and Points in Triangles
Name Example Point of Concurrency
Special Property Example
Perpendicular Bisector
CircumcenterCircumcenter is equidistant from
each vertex
Angle Bisector
IncenterIn center is
equidistant from each side
Median CentroidCentroid is two-
thirds the distance from vertex to
opposite midpoint
Altitude OrthocenterAltitudes are concurrent at orthocenter
Thursday, March 1, 2012
EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
Thursday, March 1, 2012
EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
YP =
23
YV
Thursday, March 1, 2012
EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
YP =
23
YV
YP =
23
(12)
Thursday, March 1, 2012
EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
YP =
23
YV
YP =
23
(12)
YP = 8
Thursday, March 1, 2012
EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
YP =
23
YV
YP =
23
(12)
YP = 8
PV = YV − YP
Thursday, March 1, 2012
EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
YP =
23
YV
YP =
23
(12)
YP = 8
PV = YV − YP
PV =12 − 8
Thursday, March 1, 2012
EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
YP =
23
YV
YP =
23
(12)
YP = 8
PV = YV − YP
PV =12 − 8
PV = 4
Thursday, March 1, 2012
EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
Thursday, March 1, 2012
EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
CG =
23
CE
Thursday, March 1, 2012
EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
CG =
23
CE
4 =
23
CE
Thursday, March 1, 2012
EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
CG =
23
CE
4 =
23
CE
CE = 6
Thursday, March 1, 2012
EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
CG =
23
CE
4 =
23
CE
CE = 6
GE = CE −CG
Thursday, March 1, 2012
EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
CG =
23
CE
4 =
23
CE
CE = 6
GE = CE −CG
GE = 6 − 4
Thursday, March 1, 2012
EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
CG =
23
CE
4 =
23
CE
CE = 6
GE = CE −CG
GE = 6 − 4
GE = 2
Thursday, March 1, 2012
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
EXAMPLE 3
(1, 4), (3, 0), (3, 8)
Thursday, March 1, 2012
EXAMPLE 3
x
y (1, 4), (3, 0), (3, 8)
Thursday, March 1, 2012
EXAMPLE 3
x
y (1, 4), (3, 0), (3, 8)
Thursday, March 1, 2012
EXAMPLE 3
x
y (1, 4), (3, 0), (3, 8)
Thursday, March 1, 2012
EXAMPLE 3
x
y
We need to find the centroid, so we start by finding the midpoint
of our vertical side.
(1, 4), (3, 0), (3, 8)
Thursday, March 1, 2012
EXAMPLE 3
x
y
We need to find the centroid, so we start by finding the midpoint
of our vertical side.
M =
3 + 32
,0 + 8
2
⎛⎝⎜
⎞⎠⎟
(1, 4), (3, 0), (3, 8)
Thursday, March 1, 2012
EXAMPLE 3
x
y
We need to find the centroid, so we start by finding the midpoint
of our vertical side.
M =
3 + 32
,0 + 8
2
⎛⎝⎜
⎞⎠⎟ =
62
,82
⎛⎝⎜
⎞⎠⎟
(1, 4), (3, 0), (3, 8)
Thursday, March 1, 2012
EXAMPLE 3
x
y
We need to find the centroid, so we start by finding the midpoint
of our vertical side.
M =
3 + 32
,0 + 8
2
⎛⎝⎜
⎞⎠⎟ =
62
,82
⎛⎝⎜
⎞⎠⎟
= 3,4( )
(1, 4), (3, 0), (3, 8)
Thursday, March 1, 2012
EXAMPLE 3
x
y
We need to find the centroid, so we start by finding the midpoint
of our vertical side.
M =
3 + 32
,0 + 8
2
⎛⎝⎜
⎞⎠⎟ =
62
,82
⎛⎝⎜
⎞⎠⎟
= 3,4( )
(1, 4), (3, 0), (3, 8)
Thursday, March 1, 2012
EXAMPLE 3
x
y
We need to find the centroid, so we start by finding the midpoint
of our vertical side.
M =
3 + 32
,0 + 8
2
⎛⎝⎜
⎞⎠⎟ =
62
,82
⎛⎝⎜
⎞⎠⎟
= 3,4( )
(1, 4), (3, 0), (3, 8)
Thursday, March 1, 2012
EXAMPLE 3
x
y
Thursday, March 1, 2012
EXAMPLE 3
x
y
Next, we need the distance from the opposite vertex to this
midpoint.
Thursday, March 1, 2012
EXAMPLE 3
x
y
Next, we need the distance from the opposite vertex to this
midpoint.
d = (1− 3)2 + (4 − 4)2
Thursday, March 1, 2012
EXAMPLE 3
x
y
Next, we need the distance from the opposite vertex to this
midpoint.
d = (1− 3)2 + (4 − 4)2
= (−2)2 + 02
Thursday, March 1, 2012
EXAMPLE 3
x
y
Next, we need the distance from the opposite vertex to this
midpoint.
d = (1− 3)2 + (4 − 4)2
= (−2)2 + 02
= 4
Thursday, March 1, 2012
EXAMPLE 3
x
y
Next, we need the distance from the opposite vertex to this
midpoint.
d = (1− 3)2 + (4 − 4)2
= (−2)2 + 02
= 4 = 2
Thursday, March 1, 2012
EXAMPLE 3
x
y
Thursday, March 1, 2012
EXAMPLE 3
x
y The centroid P is 2/3 of this distance from the vertex.
Thursday, March 1, 2012
EXAMPLE 3
x
y The centroid P is 2/3 of this distance from the vertex.
P = 1+
23
(2),4⎛⎝⎜
⎞⎠⎟
Thursday, March 1, 2012
EXAMPLE 3
x
y The centroid P is 2/3 of this distance from the vertex.
P = 1+
23
(2),4⎛⎝⎜
⎞⎠⎟
P = 1+
43
,4⎛⎝⎜
⎞⎠⎟
Thursday, March 1, 2012
EXAMPLE 3
x
y The centroid P is 2/3 of this distance from the vertex.
P = 1+
23
(2),4⎛⎝⎜
⎞⎠⎟
P = 1+
43
,4⎛⎝⎜
⎞⎠⎟
P =
73
,4⎛⎝⎜
⎞⎠⎟
Thursday, March 1, 2012
EXAMPLE 4The 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
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3 =
4−2
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3 =
4−2 = −2
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3 =
4−2
⊥ m =12 = −2
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3 =
4−2
⊥ m =12
m HJ( ) = 2 −1
1+ 5
= −2
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3 =
4−2
⊥ m =12
m HJ( ) = 2 −1
1+ 5 =
16
= −2
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3 =
4−2
⊥ m =12
m HJ( ) = 2 −1
1+ 5 =
16
⊥ m = −6
= −2
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
Altitude through H
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J ⊥ m =
12
, (1,2)
Altitude through H
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J ⊥ m =
12
, (1,2)
Altitude through H
y − 2 =
12
(x −1)
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J ⊥ m =
12
, (1,2)
Altitude through H
y − 2 =
12
(x −1)
y − 2 =
12
x −12
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J ⊥ m =
12
, (1,2)
Altitude through H
y − 2 =
12
(x −1)
y − 2 =
12
x −12
y =
12
x +32
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
Altitude through I
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
⊥ m = −6, (−3,−3)
Altitude through I
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
⊥ m = −6, (−3,−3)
Altitude through I
y + 3 = −6(x + 3)
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
⊥ m = −6, (−3,−3)
Altitude through I
y + 3 = −6(x + 3)
y + 3 = −6x −18
Thursday, March 1, 2012
EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
⊥ m = −6, (−3,−3)
Altitude through I
y + 3 = −6(x + 3)
y + 3 = −6x −18
y = −6x − 21
Thursday, March 1, 2012
EXAMPLE 4
Thursday, March 1, 2012
EXAMPLE 4
−6x − 21 =
12
x +32
Thursday, March 1, 2012
EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
Thursday, March 1, 2012
EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
Thursday, March 1, 2012
EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
Thursday, March 1, 2012
EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
−
452
=132
x
Thursday, March 1, 2012
EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
−
452
=132
x
213
−452
⎛⎝⎜
⎞⎠⎟=
132
x⎛⎝⎜
⎞⎠⎟
213
Thursday, March 1, 2012
EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
−
452
=132
x
213
−452
⎛⎝⎜
⎞⎠⎟=
132
x⎛⎝⎜
⎞⎠⎟
213
x = −
4513
Thursday, March 1, 2012
EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
−
452
=132
x
213
−452
⎛⎝⎜
⎞⎠⎟=
132
x⎛⎝⎜
⎞⎠⎟
213
x = −
4513
−6 −
4513
⎛⎝⎜
⎞⎠⎟− 21 =
12
−4513
⎛⎝⎜
⎞⎠⎟+
32
Thursday, March 1, 2012
EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
−
452
=132
x
213
−452
⎛⎝⎜
⎞⎠⎟=
132
x⎛⎝⎜
⎞⎠⎟
213
x = −
4513
−6 −
4513
⎛⎝⎜
⎞⎠⎟− 21 =
12
−4513
⎛⎝⎜
⎞⎠⎟+
32
−
313
= −3
13
Thursday, March 1, 2012
EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
−
452
=132
x
213
−452
⎛⎝⎜
⎞⎠⎟=
132
x⎛⎝⎜
⎞⎠⎟
213
x = −
4513
−6 −
4513
⎛⎝⎜
⎞⎠⎟− 21 =
12
−4513
⎛⎝⎜
⎞⎠⎟+
32
−
313
= −3
13
−
4513
,−3
13
⎛⎝⎜
⎞⎠⎟
Thursday, March 1, 2012
CHECK YOUR UNDERSTANDING
p. 337 #1-4
Thursday, March 1, 2012
PROBLEM SET
Thursday, March 1, 2012
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