triangle centres

Triangle Centres

Mental Health Break

Given the following triangle, find the:

centroid

orthocenter

circumcenter

4,1A

2,1 B

1,5C

4,1A

2,1 B

1,5C

D

EF

Equation of AD (median)

Strategy….

1. Find midpoint D

2. Find eq’n of AD by

- Find slope “m” of AD using A & D

- Plug “m” & point A or D into y=mx+b & solve for “b”

- Now write eq’n using “m” & “b”

Remember – the centroid is useful as the centre of the mass of a triangle – you can balance a triangle on a centroid!

4,1A

2,1 B

1,5C

D

EF

Equation of AD (median)

4,1A

2,1 B

1,5C

D

EF

Equation of BE (median)

Strategy….

1. Find midpoint E

2. Find eq’n of BE by

- Find slope “m” of BE using B & E

- Plug “m” & point B or E into y=mx+b & solve for “b”

- Now write eq’n using “m” & “b”

4,1A

2,1 B

1,5C

D

EF

Equation of BE (median)

4,1A

2,1 B

1,5C

D

EF

Question?

Do we have to find the equation of median CF also?

4,1A

2,1 B

1,5C

D

EF

No

We only need the equations of 2 medians…

So, what do we do now?

4,1A

2,1 B

1,5C

D

EF

We need to find the Point of Intersection for medians AD & BE using either substitution or elimination

4,1A

2,1 B

1,5C

D

EF

Equation of median AD

Equation of median BE

4,1A

2,1 B

1,5C

D

EF

Equation of AD (median)

Strategy….

1. Find midpoint D

2. Find eq’n of AD by

- Find slope “m” of AD using A & D

- Plug “m” & point A or D into y=mx+b & solve for “b”

- Now write eq’n using “m” & “b”

Centroid Eq’n AD – Midpoint of BC

)2

1,2(

)2

1,

2

4(

)2

12,

2

51(

DM

Centroid Eq’n AD – Slope of AD

2

36

93

1

2

9329

)1(2

421

12

12

x

xx

yymAD

Centroid Eq’n AD – Finding “b”

2

52

3

2

82

34

)1)(2

3(4

b

b

b

b

bmxy

Centroid Eq’n AD – Equation

2

5

2

3

xy

bmxy

Centroid Eq’n BE – Midpoint of AC

)2

5,2(

)2

5,

2

4(

)2

14,

2

51(

EM

Centroid Eq’n BE – Slope of BE

2

36

93

1

2

9329

)1(2

)2(25

12

12

x

xx

yymAD

Centroid Eq’n BE – Finding “b”

2

12

3

2

42

3

2

4

)1)(2

3(2

b

b

b

b

bmxy

Centroid Eq’n BE – Equation

2

1

2

3

xy

bmxy

Centroid – Intersection of Eq’n AD & BE

2

1

2

3 xy

EquationBE

2

5

2

3 xy

EquationAD

Centroid – Intersection of Eq’n AD & BE

2

1

2

3 xy BE

2

5

2

3 xy AD

2

42 y Add AD and BE

1

22

y

y Simplify and solve for y

Centroid – Intersection of Eq’n AD & BE

Substitute y = 1 into one of the equations

12

3

2

32

3

2

1

2

22

1

2

3

2

22

1

2

31

x

x

x

x

x

Therefore, the point of intersection is (1,1)

4,1A

2,1 B

1,5C

D

EF

Equation of altitude AD

Strategy….

1. Find “m” of BC

2. Take –ve reciprocal of “m” of BC to get “m” of AD

3. Find eq’n of AD by

- Plug “m” from 2. & point A into y=mx+b & solve for “b”

- Now write eq’n using “m” & “b”

Centroid – Intersection of Eq’n AD & BE

Therefore, the Centroid is (1,1)

4,1A

2,1 B

1,5C

D

EF

Equation of altitude AD

4,1A

2,1 B

1,5C

D

EF

Equation of altitude BE

Strategy….

1. Find “m” of AC

2. Take –ve reciprocal of “m” of AC to get “m” of BE

3. Find eq’n of BE by

- Plug “m” from 2. & point B into y=mx+b & solve for “b”

- Now write eq’n using “m” & “b”

4,1A

2,1 B

1,5C

D

EF

Equation of altitude BE

4,1A

2,1 B

1,5C

D

EF

Question?

Do we have to find the equation of altitude CF also?

4,1A

2,1 B

1,5C

D

EF

No

We only need the equations of 2 altitudes…

So, what do we do now?

4,1A

2,1 B

1,5C

D

EF

We need to find the Point of Intersection for altitudes AD & BE using either substitution or elimination

4,1A

2,1 B

1,5C

D

EF

Equation of altitude AD

Equation of altitude BE

Orthocentre Eq’n AD – Slope of BC then Slope of AD

reciprocal

negative

m

and

xx

yym

AD

AD

2

2

16

3

)1(5

)2(112

12

Orthocentre Eq’n AD – Finding “b”

2

24

)1)(2(4

b

b

b

bmxy

Orthocentre Eq’n AD – Equation

22 xy

bmxy

Orthocentre Eq’n BE – Slope of AC then Slope of BE

reciprocal

negative

m

and

xx

yym

BE

AC

2

2

16

3

)1(5

4112

12

Orthocentre Eq’n BE – Finding “b”

0

22

)1)(2(2

b

b

b

bmxy

Orthocentre Eq’n BE – Equation

xy

xy

bmxy

2

02

Orthocentre – Intersection of Eq’n AD & BE

xy

EquationBE

222 xy

EquationAD

Orthocentre – Intersection of Eq’n AD & BE

xy 2 BE

22 xy AD

22 y Add AD and BE

1y Simplify and solve for y

Orthocentre – Intersection of Eq’n AD & BE

Substitute y = 1 into one of the equations

2

12

2

2

1

21

2

x

x

x

xyTherefore, the point of intersection or Orthocentre

)1,2

1(

A

B

C

D

E

F Equation of ED (perpendicular bisector)Strategy… (use A (-1, 4), B (-1, -2) & C(5, 1))

1. Find midpoint D

2. Find eq’n of ED by- Find slope “m” of BC using

B & E

- Take –ve reciprocal to get “m” of ED

- Plug “m” ED & point D into y = mx+b & solve for b

- Now write eq’n using “m” & “b”

G

IH

A

B

C

D

E

FEquation of ED (perpendicular bisector)

G

IH

A

B

C

D

E

F Equation of FG (perpendicular bisector)Strategy… (use A (-1, 4), B (-1, -2) & C(5, 1))

1. Find midpoint F

2. Find eq’n of ED by- Find slope “m” of AC using

A & C

- Take –ve reciprocal to get “m” of FG

- Plug “m” FG & point F into y = mx+b & solve for b

- Now write eq’n using “m” & “b”

G

IH

A

B

C

D

E

FQuestion?

Do we have to find the equation of perpendicular bisector HI?

G

IH

A

B

C

D

E

FNo

We only need the equations of 2 perpendicular bisectors…

So, what do we do now?

G

IH

A

B

C

D

E

FWe need to find the Point of Intersection for perpendicular bisectors ED & FG using either substitution or elimination

G

IH

A

B

C

D

E

FEquation of perpendicular bisector ED

Equation of perpendicular bisector FG

G

IH

A

B

C

D

E

FEquation of FG (perpendicular bisector)

G

IH

Circumcentre eq’n ED – Midpoint of BC

)2

1,2(

)2

1,

2

4(

)2

12,

2

51(

DM

Circumcentre Eq’n ED – Slope of BC & Slope of ED

2

2

16

3

)1(5

)2(112

12

ED

BC

m

So

xx

yym

Circumcentre Eq’n ED – Finding “b”

2

72

8

2

1

42

1

)2)(2(2

1

b

b

b

b

bmxy

Circumcentre Eq’n ED – Equation

2

72

xy

bmxy

Circumcentre Eq’n FG – Midpoint of AC

)2

5,2(

)2

5,

2

4(

)2

14,

2

51(

EM

Circumcentre Eq’n FG – Slope of AC & Slope of FG

2

2

16

3

)1(5

4112

12

FG

AC

m

So

xx

yym

Circumcentre Eq’n FG – Finding “b”

2

32

8

2

5

42

5

)2)(2(2

5

b

b

b

b

bmxy

Circumcentre Eq’n FG – Equation

2

32

xy

bmxy

Circumcentre – Intersection of Eq’n ED & FG

2

32 xy

EquationFG

2

72 xy

EquationED

Circumcentre – Intersection of Eq’n ED & FG

2

32 xy

FG

2

72 xy ED

12

4

2

2

222

42

y

y

y

y

Add ED and FG

Simplify and solve for y

Orthocentre – Intersection of Eq’n AD & BE

Substitute y = 1 into one of the equations

4

5

)2

1)(

2

5(

2

2

225

22

7

2

22

721

2

72

x

x

x

x

x

xy

Therefore, the point of intersection or Circumcentre

)1,4

5(