M911 LES27

892019 M911 LES27

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ANALYTICAL GEOMETRY 9830802983081

Learning Outcomes and Assessment Standards

Learning Outcome 3 Space shape and measurement

Assessment Standard AS 3(c) and AS 3(a)

The gradient and inclination of a straight line

The equation of a straight line

bull

bull

Overview

In this lesson you will

Discover what is meant by inclination

Use trigonometry to find the inclination of a straight line

Find the angle between two straight lines

Use analytical methods to find the three angles of a triangle

Lesson

The inclination of a straight line

Definition The angle formed by a straight line and the positive direction of the

horizontal

4 OPTIONS

To find this angle we need the concept of a gradient and link it directly to

trigonometry

m =change in y

__

change in x

and tan α = y

_ x

there4 For any line segment AB

tan α = mAB

there4 tan α = y

B ndash y

A

_

x

B

ndash x

A

So α = tanndash1 (mAB

)

y

x

y

x

27LESSON

Ov er v iew

a

B ( x B y

B)

A ( x A yA) θ

A (1 4)

B (ndash3 ndash2)

Lesson

892019 M911 LES27


Lesson 1 | Algebr a

P age 1P ag e 115

Examples

1 We introduce any horizontal line through AB

Then tan q = mAB

= 4 + 2

_

1 + 3

= 6

_ 4

=3

_ 2

there4 q = tanndash1 ( 3

_ 2

) = 563deg

2 tan q = mPQ

= 6 + 4

_

ndash2 ndash 1

= ndash 10

_ 3

there4 q = tanndash1 ( ndash 10

_ 3

) = ndash733deg

Notice that when tan q lt 0 we are working with the negative angle q So to

get to q we need to add a period of tan which is 180deg

So q = qprime + 180deg = ndash733deg + 180deg = 1067deg

3 Find the size of the angles between the following lines

a y = ndash x + 4 and y = 1

_ 2

x + 3

b y = ndash3 x ndash 4 and y = ndash x

c y = 8 x and y = 2 x + 3

a For L1 tan a

2 lt 0

So a2 = tanndash1(ndash1) + 180deg

= ndash45deg + 180deg

= 135deg

a2prime = 45deg

For L2 a

1 tanndash1 ( 1

_ 2

) + 266deg

The acute angle between the lines will be 45deg + 266deg = 716deg

The obtuse angle between them is 180deg ndash 716deg = 1084deg

b For L1 m = ndash1

q1 = tanndash1(ndash1) + 180deg


there4 q1 = 135deg

and q1prime = 45deg

For L2 m = ndash3



q2 = 1084deg

Thus obtuse angle 1084deg + 45 = 1534deg

θ

θrsquo

Q(ndash2 6)

P(1 ndash4)

θ

θrsquo

Q(ndash2 6)

P(1 ndash4)

To use this concept

effectively always

draw a diagram

L1

L2

y = ndash3 x ndash 4

y = ndash x

q1

q1

q2

a1

a2

a2

L1L

2

y = x + 3

y = ndash x + 4

1

2

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Acute angle 266deg

c L1 m = 8 ndash q

1 = tanndash1(8) = 828deg

L2 m = 2 ndash q


there4 ang between lines (acute) = 828 ndash 634deg

= 194deg

The obtuse angle will be = 180deg ndash 194deg

= 1606deg

4 A(ndash1 5) B(2 3) and C(ndash6 ndash1) are co-ordinates of the vertices of ABC

Find the size of the angles

Draw a picture

Tip Find α β and θ the use geometry

mAC

= 6

_ 5

mBC

= 4

_ 8

mAB

= 2

_

ndash3

tan α = 6

_ 5

tan β = 1

_ 2

tan θ = ndash 2

_ 3

α = 502deg β = 266deg θ = 1463deg

By geometry ^ C = α ndash β

^ C = 236deg

^ A = θ ndash α

^ A = 961deg

^ B1 = 603 (ltrsquos in )

Letrsquos find the inclination of each side

First

AC m =5 + 1

_

ndash1 + 6

=6

_

5


_ 5

) = 502deg

AB m = 5 ndash 3

_

ndash1 ndash 2 = ndash 2

_ 3

there4 b = tanndash1 ( ndash 2

_ 3

) + 180deg

= 1463deg

BC m = 3 + 1

_

2 + 6 = 1

_ 2

a = tanndash1 ( ndash 1

_ 2

) = 266deg

^ C = q ndash a = 502deg 266deg

^ C = 236deg

A(ndash1 5)

B(2 3

C(ndash6 ndash1)

βα

θ

q1

q2

L2

L1

y = 2 x + 3

y = 8 x

1463deg

A(ndash1 5)

B(2 3)

502deg

266deg

266deg

E xample

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6 ^ A = 45deg Find the gradient of ℓ2

7 Find the gradient of AB

8 Calculate the gradient of OC

9 P(ndash1 4) Q(2 2) and

R(ndash6 ndash1) are co-ordinates of PQR find the angles of the triangle

892019 M911 LES27


Lesson 1 | Algebr a

P age 1P ag e 115

Examples

1 We introduce any horizontal line through AB

Then tan q = mAB

= 4 + 2

_

1 + 3

= 6

_ 4

=3

_ 2


_ 2

) = 563deg

2 tan q = mPQ

= 6 + 4

_

ndash2 ndash 1

= ndash 10

_ 3

there4 q = tanndash1 ( ndash 10

_ 3

) = ndash733deg

Notice that when tan q lt 0 we are working with the negative angle q So to

get to q we need to add a period of tan which is 180deg

So q = qprime + 180deg = ndash733deg + 180deg = 1067deg

3 Find the size of the angles between the following lines

a y = ndash x + 4 and y = 1

_ 2

x + 3

b y = ndash3 x ndash 4 and y = ndash x

c y = 8 x and y = 2 x + 3

a For L1 tan a

2 lt 0

So a2 = tanndash1(ndash1) + 180deg


= 135deg

a2prime = 45deg

For L2 a

1 tanndash1 ( 1

_ 2

) + 266deg

The acute angle between the lines will be 45deg + 266deg = 716deg

The obtuse angle between them is 180deg ndash 716deg = 1084deg

b For L1 m = ndash1



there4 q1 = 135deg

and q1prime = 45deg

For L2 m = ndash3



q2 = 1084deg

Thus obtuse angle 1084deg + 45 = 1534deg

θ

θrsquo

Q(ndash2 6)

P(1 ndash4)

θ

θrsquo

Q(ndash2 6)

P(1 ndash4)

To use this concept

effectively always

draw a diagram

L1

L2

y = ndash3 x ndash 4

y = ndash x

q1

q1

q2

a1

a2

a2

L1L

2

y = x + 3

y = ndash x + 4

1

2

892019 M911 LES27


ge 116

Acute angle 266deg

c L1 m = 8 ndash q


L2 m = 2 ndash q



= 194deg


= 1606deg



Draw a picture


mAC

= 6

_ 5

mBC

= 4

_ 8

mAB

= 2

_

ndash3

tan α = 6

_ 5

tan β = 1

_ 2

tan θ = ndash 2

_ 3

α = 502deg β = 266deg θ = 1463deg


^ C = 236deg

^ A = θ ndash α

^ A = 961deg



First

AC m =5 + 1

_

ndash1 + 6

=6

_

5


_ 5

) = 502deg

AB m = 5 ndash 3

_


_ 3


_ 3

) + 180deg

= 1463deg

BC m = 3 + 1

_

2 + 6 = 1

_ 2


_ 2

) = 266deg


^ C = 236deg

A(ndash1 5)

B(2 3

C(ndash6 ndash1)

βα

θ

q1

q2

L2

L1

y = 2 x + 3

y = 8 x

1463deg

A(ndash1 5)

B(2 3)

502deg

266deg

266deg

E xample

892019 M911 LES27


892019 M911 LES27


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892019 M911 LES27


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Acute angle 266deg

c L1 m = 8 ndash q


L2 m = 2 ndash q



= 194deg


= 1606deg



Draw a picture


mAC

= 6

_ 5

mBC

= 4

_ 8

mAB

= 2

_

ndash3

tan α = 6

_ 5

tan β = 1

_ 2

tan θ = ndash 2

_ 3

α = 502deg β = 266deg θ = 1463deg


^ C = 236deg

^ A = θ ndash α

^ A = 961deg



First

AC m =5 + 1

_

ndash1 + 6

=6

_

5


_ 5

) = 502deg

AB m = 5 ndash 3

_


_ 3


_ 3

) + 180deg

= 1463deg

BC m = 3 + 1

_

2 + 6 = 1

_ 2


_ 2

) = 266deg


^ C = 236deg

A(ndash1 5)

B(2 3

C(ndash6 ndash1)

βα

θ

q1

q2

L2

L1

y = 2 x + 3

y = 8 x

1463deg

A(ndash1 5)

B(2 3)

502deg

266deg

266deg

E xample

892019 M911 LES27


892019 M911 LES27


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892019 M911 LES27


892019 M911 LES27


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892019 M911 LES27


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M911 LES27