CONFIDENTIAL1 Geometry Medians and Altitudes of Triangle

CONFIDENTIAL 1

Geometry

Medians and Altitudes of

Triangle

CONFIDENTIAL 2

Warm up

1) NQ, QP, and QM are perpendicular bisectors of JKL. Find each Measure.

a) KL b) QJ c) m JQL

K

PN

M

36˚

40˚ 9.1

L

Q

7

J

CONFIDENTIAL 3

A median of a triangle is a segment whose endpoints are a vertex of the triangle and

the midpoint of the opposite side.

Medians and Altitudes of Triangle

median

A B

C

DEvery triangle has three medians, and the median are

concurrent, as shown in the next slide.

CONFIDENTIAL 4

Construction Centroid of a Triangle

A

B

C

X

Z

Y

A

B

C

X

Z

Y

Next page :

Draw ABC. Construct themidpoint of AB, BC, and AC.Label the midpoints of the sidesX, Y, and Z, respectively.

Darw AY, BZ, and CX. Theseare the three median of ABC.

CONFIDENTIAL 5

A

B

C

X

Z

YP

Label the point where AY, BZ,and CX intersect as P.

CONFIDENTIAL 6

The point of concurrency of the medians of a triangle is the centroid of the triangle. The centroid is always inside the triangle. The

centroid is also called the center of gravity because it is the point where a triangular region

will balance.

CONFIDENTIAL 7

Theorem 3.1 Centroid Theorem

The centroid of a triangle is located 2 of the distance from 3

each vertex to the midpoint of the opposite side.

AP = 2

3AY BP =

2

3BZ CP =

2

3CX

A

B

C

X

Z

YP

CONFIDENTIAL 8

A

B

C

E

D

F

G

Using the Centroid to Find Segment Lengths

In ∆ABC, AF = 9, and GE = 2.4. Find each length.

A) AG

Centroid Thm.

Substitute 9 for AF

Simplify.

AG = 2

3AF

AG = 2

3(9)

AG = 6

Next page :

CONFIDENTIAL 9

B) CE

CG = 2

3CE

CG + GE = CE

2

3CE + GE = CE

GE = 1

3CE

2.4 = 1

3CE

7.2 = CE

Centroid Thm.Seg. Add. Post.Substitute 2 CE for CG. 3Subtract 2CE from both sides. 3

Substitute 2.4 for GE.Multiply both sides by 3.

A

B

C

E

D

F

G

CONFIDENTIAL 10

Now you try!

1) In ∆JKL, ZW =7,and LX = 8.1.

Find each length.

a) KW

b) LZ W

K

Z

L

JX

CONFIDENTIAL 11

R(6, 4)

y

xP(3, 0)

Q(0, 8)

0

2

4

6

8

2 4 6 8

Problem-Solving Application

The diagram shows the plan for a triangular piece of a mobile. Where should the sculptor attach the support so that the triangle is balanced?

1) Understand the Problem

The answer will be the coordinates of the centroid of ∆PQR. The important information is the location of the vertices, P(3,0),Q(0,8), and R(6,4).

Next page :

CONFIDENTIAL 12

R(6, 4)

y

xP(3, 0)

Q(0, 8)

0

2

4

6

8

2 4 6 8

2) Make a Plan

The centroid of the triangle is the point of intersection of the three medians. So write the equations for two medians and find their point of intersection.

CONFIDENTIAL 13

R(6, 4)

y

xP(3, 0)

Q(0, 8)

0

2

4

6

8

2 4 6 8

Let M be the midpoint of QR and N Be the midpoint of QP.

m = 0+6

2,8+4

2 = (3, 6) n =

0+3

2,8+0

2 = (1.5, 4)

PM is vertical. Its equation is x = 3. RN is horizontal.Its equation is y =4. The coordinates of the centroid are S(3, 4).

CONFIDENTIAL 14

Now you try!

2) Find the average of the x-coordinates and the average of the y-coordinates of the vertices of ∆PQR. Make a conjecture

about the centroid of a triangle.

R(6, 4)

y

xP(3, 0)

Q(0, 8)

0

2

4

6

8

2 4 6 8

CONFIDENTIAL 15

An altitude of a triangle is a perpendicular segment from a vertex to the line containing the opposite side. Every triangle has three altitudes. An altitude can be inside, outside, or on the triangle.

In ∆QRS, altitude QY is inside the

triangle, but RX and SZ are not.

Notice that the lines containing the

altitudes are concurrent at P. This

point of concurrency is the

orthocenter of the triangle.

Q

R

SP Z

X

Y

CONFIDENTIAL 16

Finding the Orthocenter

Next page :

(-2, 4)

K

LJ

X = -2

x

y

0-4Y = x + 6

7

2

Find the orthocenter of ∆JKL with vertices J(-4,2), K(-2,6), and L(2,2).

Step 1 Graph the triangle.

Step 2 Find an equation of the line containing the altitude from K to JL.

Since JL is horizontal, the altitude is vertical. The line containing it must pass through K(-2,6), so the equation of the line is x = -2.

CONFIDENTIAL 17

Next page :

(-2, 4)

K

LJ

X = -2

x

y

0-4Y = x + 6

7

2

Step 3 Find an equation of the line containing the altitude from J to KL.

slope of KL = 2 - 6

2- (-2) = -1

The slope of a line

perpendicular to KL is 1.

This line must pass

through J(-4, 2).

CONFIDENTIAL 18

(-2, 4)

K

LJ

X = -2

x

y

0-4Y = x + 6

7

2

y - y1 = m(x - x1)

y - 2 = 1[x-(-4)]

y - 2 = x + 4

y = x + 6

Point-slope form

Substitute 2 for y1. 1 for m, and -4 for x1.

Distribute 1.

Add 2 to both sides.

Step 4 Solve the system to find the coordinates of the orthocenter.

x = -2

y = x + 2

y = -2 + 6 = 4

The coordinates of the orthocenter are (-2,4).

Substitute -2for x

CONFIDENTIAL 19

Now you try!

3) Show that the altitude to JK passes through the orthocenter of ∆JKL.

(-2, 4)

K

LJ

X = -2

x

y

0-4Y = x + 6

7

2

CONFIDENTIAL 20

Now some problems for you to practice !

CONFIDENTIAL 21

VX = 205, and RW = 104. Find each length.

1) VW 2) WX

3) RY 4) WY

WY

X

R

V

T

Assessment

CONFIDENTIAL 22

0

4

2

y

x

B(7, 4)

A(0, 2)

2 C(5, 0) 8

5) The diagram shows a plan for a piece of a mobile. A chain will hang from the centroid of the triangle. At what coordinates

should the artist attach the chain?

CONFIDENTIAL 23

Find the orthocenter of a triangle with the given vertices.

6) K(2, -2), L(4, 6), M(8, -2)

7) U(-4, -9), V(-4, 6),W(5, -3)

8) P(-5, 8), Q(4, 5), R(-2, 5)

9) C(-1, -3), D(-1, 2), E(9, 2)

CONFIDENTIAL 24

Let’s review

A median of a triangle is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side.

Medians and Altitudes of Triangle

median

A B

C

DEvery triangle has three medians, and the median are concurrent, as shown in the construction below.

CONFIDENTIAL 25

Construction Centroid of a Triangle

A

B

C

X

Z

Y

A

B

C

X

Z

Y

Next page :

Draw ABC. Construct themidpoint of AB, BC, and AC.Label the midpoints of the sidesX, Y, and Z, respectively.

Darw AY, BZ, and CX. Theseare the three median of ABC.

CONFIDENTIAL 26

A

B

C

X

Z

YP

Label the point where AY, BZ,and CX intersect as P.

CONFIDENTIAL 27

The point of concurrency of the medians of a triangle is the centroid of the triangle. The centroid is always inside the triangle. The

centroid is also called the center of gravity because it is the point where a triangular region

will balance.

CONFIDENTIAL 28

Theorem 3.1 Centroid Theorem

The centroid of a triangle is located 2/3 of the distance from each vertex to the midpoint of the opposite side.

AP = 2

3AY BP =

2

3BZ CP =

2

3CX

A

B

C

X

Z

YP

CONFIDENTIAL 29

A

B

C

E

D

F

G

Using the Centroid to Find Segment Lengths

In ∆ABC, AF = 9, and GE = 2.4. Find each length.

A) AG

Centroid Thm.

Substitute 9 for AF

Simplify.

AG = 2

3AF

AG = 2

3(9)

AG = 6

Next page :

CONFIDENTIAL 30

B) CE

CG = 2

3CE

CG + GE = CE

2

3CE + GE = CE

GE = 1

3CE

2.4 = 1

3CE

7.2 = CE

Centroid Thm.

Seg. Add. Post.

Substitute 2/3 CE for CG.

Subtract 2/3 CE from both sides.

Substitute 2.4 for GE.

Multiply both sides by 3.

A

B

E

D

G

CONFIDENTIAL 31

R(6, 4)

y

xP(3, 0)

Q(0, 8)

0

2

4

6

8

2 4 6 8

Problem-Solving Application

The diagram shows the plan for a triangular piece of a mobile. Where should the sculptor attach the support so that the triangle is balanced?

1) Understand the Problem

The answer will be the coordinates of the centroid of ∆PQR. The important information is the location of the vertices, P(3,0),Q(0,8), and R(6,4).

Next page :

CONFIDENTIAL 32

R(6, 4)

y

xP(3, 0)

Q(0, 8)

0

2

4

6

8

2 4 6 8

2) Make a Plan

The centroid of the triangle is the point of intersection of the three medians. So write the equations for two medians and find their point of intersection.

CONFIDENTIAL 33

R(6, 4)

y

xP(3, 0)

Q(0, 8)

0

2

4

6

8

2 4 6 8

Let M be the midpoint of QR and N Be the midpoint of QP.

m = 0+6

2,8+4

2 = (3, 6) n =

0+3

2,8+0

2 = (1.5, 4)

PM is vertical. Its equation is x = 3. RN is horizontal.Its equation is y =4. The coordinates of the centroid are S(3, 4).

CONFIDENTIAL 34

An altitude of a triangle is a perpendicular segment from a vertex to the line containing the opposite side. Every triangle has three altitudes. An altitude can be inside, outside, or on the triangle.

In ∆QRS, altitude QY is inside the

triangle, but RX and SZ are not.

Notice that the lines containing the

altitudes are concurrent at P. This

point of concurrency is the

orthocenter of the triangle.

Q

R

SP Z

X

Y

CONFIDENTIAL 35

Finding the Orthocenter

Next page :

(-2, 4)

K

LJ

X = -2

x

y

0-4Y = x + 6

7

2

Find the orthocenter of ∆JKL with vertices J(-4,2), K(-2,6), and L(2,2).

Step 1 Graph the triangle.

Step 2 Find an equation of the line containing the altitude from K to JL.

Since JL is horizontal, the altitude is vertical. The line containing it must pass through K(-2,6), so the equation of the line is x = -2.

CONFIDENTIAL 36

Next page :

(-2, 4)

K

LJ

X = -2

x

y

0-4Y = x + 6

7

2

Step 3 Find an equation of the line containing the altitude from J to KL.

slope of KL = 2 - 6

2- (-2) = -1

The slope of a line

perpendicular to KL is 1.

This line must pass

through J(-4, 2).

CONFIDENTIAL 37

(-2, 4)

K

LJ

X = -2

x

y

0-4Y = x + 6

7

2

y - y1 = m(x - x1)

y - 2 = 1[x-(-4)]

y - 2 = x + 4

y = x + 6

Point-slope form

Substitute 2 for y1. 1 for m, and -4 for x1.

Distribute 1.

Add 2 to both sides.

Step 4 Solve the system to find the coordinates of the orthocenter.

x = -2

y = x + 2

y = -2 + 6 = 4

The coordinates of the orthocenter are (-2,4).

Substitute -2for x

CONFIDENTIAL 38

You did a great job today!