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Geometry Section 5-2 1112

The document discusses medians and altitudes of triangles. It defines key terms like median, centroid, altitude and orthocenter. A median connects a vertex to the midpoint of the opposite side, while an altitude connects a vertex perpendicular to the opposite side. The centroid is the point where the medians intersect and is always inside the triangle. The orthocenter is the point where the altitudes intersect. Examples are provided to demonstrate calculating lengths and finding points related to medians, altitudes, centroids and orthocenters using given information about triangles.

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SECTION 5-2 Medians 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
VOCABULARY 1. Median: 2. Centroid: 3. Altitude: 4. Orthocenter: Thursday, March 1, 2012
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
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
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
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
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 2 2 2 AG = AF , BG = BD, and CG = CE 3 3 3 Thursday, March 1, 2012
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
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. 2 YP = YV 3 Thursday, March 1, 2012
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
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
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
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
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
EXAMPLE 2 In ∆ABC, CG = 4. Find GE. Thursday, March 1, 2012
EXAMPLE 2 In ∆ABC, CG = 4. Find GE. 2 CG = CE 3 Thursday, March 1, 2012
EXAMPLE 2 In ∆ABC, CG = 4. Find GE. 2 CG = CE 3 2 4 = CE 3 Thursday, March 1, 2012
EXAMPLE 2 In ∆ABC, CG = 4. Find GE. 2 CG = CE 3 2 4 = CE 3 CE = 6 Thursday, March 1, 2012
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
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
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
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 y (1, 4), (3, 0), (3, 8) x Thursday, March 1, 2012
EXAMPLE 3 y (1, 4), (3, 0), (3, 8) x Thursday, March 1, 2012
EXAMPLE 3 y (1, 4), (3, 0), (3, 8) x Thursday, March 1, 2012
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
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
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
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
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
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
EXAMPLE 3 y x Thursday, March 1, 2012
EXAMPLE 3 y Next, we need the distance from the opposite vertex to this midpoint. x Thursday, March 1, 2012
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
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
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
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
EXAMPLE 3 y x Thursday, March 1, 2012
EXAMPLE 3 y The centroid P is 2/3 of this distance from the vertex. x Thursday, March 1, 2012
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
EXAMPLE 4 Thursday, March 1, 2012
EXAMPLE 4 1 3 −6x − 21 = x + 2 2 Thursday, March 1, 2012
EXAMPLE 4 1 3 −6x − 21 = x + 2 2 +6x +6x Thursday, March 1, 2012
EXAMPLE 4 1 3 −6x − 21 = x + 2 2 +6x +6x 13 3 −21 = x + 2 2 Thursday, March 1, 2012
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
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
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
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
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
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
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
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

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Geometry Section 5-2 1112

  • 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 - CENTROIDTHEOREM 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 - CENTROIDTHEOREM 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 andPoints 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.
  • 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 Thursday, March 1, 2012
  • 83.
  • 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