triangle centres
DESCRIPTION
Triangle Centres. Mental Health Break. Given the following triangle, find the: centroid orthocenter circumcenter. Centroid. Equation of AD (median) Strategy…. Find midpoint D 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” - PowerPoint PPT PresentationTRANSCRIPT
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(