HIGHER MATHEMATICS

Unit 1 – Topic 1

The Straight Line

THE DISTANCE FORMULA

A is the point (x1 , y1) and B is the point (x2 , y2)

By Pythagoras’ Theorem:

AB2 = AC2 + CB2

AB2 = (x2 – x1)2 + (y2 – y1)

2

AB = [(x2 – x1)2 + (y2 – y1)

2]

So the DISTANCE formula is:

AB = [(x2 – x1)2 + (y2 – y1)

2]

A

By

xO

THE MIDPOINT FORMULA

A is the point (x1 , y1) and B is the point (x2 , y2)

The Midpoint, M, is given by:

( )A

By

xO

Mx1 + x2

2

y1 + y2

2,

EXAMPLE 1 a) Calculate the length of the line joining pts A(3 , -4) & B(-5 , 7).

AB = [(x2 – x1)2 + (y2 – y1)

2]

AB = [(-5 – 3)2 + (7 – (-4))2]

AB = (64 + 121)

AB = (185)AB = 13.6 units

b) Find the midpoint of the line above.

( )x1 + x2

2

y1 + y2

2M = ,

( )3 + (-5)

2M = , -4 + 7

2

M = (-1 , 1.5)

EXAMPLE 2 Prove that the triangle with verticesA(-7 , 9), B(3 , 13) and C(2 , 1)is isosceles.

AC = [(x2 – x1)2 + (y2 – y1)

2]

AC = [(2 – (-7))2 + (1 – 9)2]

AC = (81 + 64)

AC = 145

AB = [(3 – (-7))2 + (13 – 9)2]

AB = (100 + 16)

AB = 116

BC = [(2 – 3)2 + (1 – 13)2]

BC = (1 + 144)

BC = 145

SinceAC = BC,

triangle ABC is isosceles.

Exercises from MIA book:

Page 2 Ex 1 Qu 3 - 13

Exercises from Heinemann book:

Page ? Ex ? Qu ??

GRADIENT OF A STRAIGHT LINE

Gradient =Change in y

Change in x

mAB = y2 – y1

x2 – x1

Also,tan =

Opposite

Adjacent

y2 – y1

x2 – x1

=mAB

A is the point (x1 , y1) and B is the point (x2 , y2)

A

By

xO

So, mAB = tan

mAB =y2 – y1

x2 – x1

mAB = tan

where is the angle between the line and the positive direction of the x-axis.

If lines are Parallel then they have equal gradients!If lines have equal gradients then they are parallel!!

Points that lie on the same straight line are said to be COLLINEAR.

To prove COLLINEARITY show that they have a common point and that they have equal gradients, i.e. PARALLEL!

x

Line slopesUP fromLEFT to RIGHT

tan is POSITIVE

GRADIENTis

POSITIVE

Line isHORIZONTAL

= 0

tan = 0

GRADIENT is

ZERO

Line slopesDOWN from

LEFT to RIGHT

tan is NEGATIVE

GRADIENTis

NEGATIVE

Line is VERTICAL

= 90

tan NOT DEFINED

GRADIENT is NOT

DEFINED

mAB = tanmPQ =

y2 – y1

x2 – x1

8 – (-7)

6 – (–4)=

= 15

10=

= tan

= tan-1

EXAMPLE 3 Calculate the gradient of the line joining points P(-4 , -7) & Q(6 , 8) and calculate the angle the line makes with the x-axis.

3

2

3

23

2

= 56.3º

mAB =y2 – y1

x2 – x1

3 – 6

-1 – 8=

= -3

-9=

EXAMPLE 4 Show that points A(8 , 6), B(–1 , 3) and C(-4 , 2) are COLLINEAR.

1

3

mBC =y2 – y1

x2 – x1

2 – 3

-4 –(-1)=

= -1

-3= 1

3

Since mAB = mBC AB is parallel to BC.Also, B is a common point so A, B and C

are Collinear.

Exercises from MIA book:

Page 4 Ex 2A/B All Qu.

Exercises from Heinemann book:

Page ? Ex ? Qu ??

PERPENDICULAR LINES

O x

y

A(a , b)B(-b , a)

Line OA is rotated 90o

anti-clockwise to OB.

mOA =b - 0

a - 0

b

a=

mOB =a - 0

-b - 0

a

-b=

Notice: mOA x mOB

b

a=

a

-bx = -1

If m1 x m2 =-1,then the lines arePERPENDICULAR.

If the lines arePERPENDICULAR,then m1 x m2 =-1

m1 m2

2

¼

-¾

-½

-4

43

m1 and m2 are the RECIPROCALS of the other

(with the sign changed).

You need only remember this simple rule:

turn upside down andchange the sign.

EXAMPLE 5 Write down the gradient of the lines PERPENDICULAR to the following:

a) y = 8x + 5

m = 8

mperp = -⅛

b) y = -4x - 7

m = -4

mperp = ¼

c) 6y = 5x – 12

m = 5/6

mperp = -6/5

y = 5/6x - 2

EXAMPLE 6 Prove that the triangle with vertices at A(4 , 1), B(7 , 5) and C(0 , 4) is right-angled.

mAB =y2 – y1

x2 – x1

5 – 1

7 – 4=

=4

3

mAC =4 – 1

0 – 4

=3

-4

Since mAB x mAC =-1,then the lines arePERPENDICULAR.

So ∆ABC must beright-angled at A

Exercises from MIA book:

Page 7 Ex 3 All Qu.

Exercises from Heinemann book:

Page ? Ex ? Qu ??

EQUATION OF A STRAIGHT LINE

A is the point (0 , c) and B is the point (x , y)

A

By

xO

mAB =y2 – y1

x2 – x1

y – c

x – 0=

m = y - c

xmx = y - c

So we get:

y = mx + c

y-InterceptGradient

a) m = –5 through Pt(0 , 4) y = -5x + 4

EXAMPLE 7 Write down the equations below:

b) m = 3 through Pt(0 , -2) y = 3x - 2

c) m = -1 through Pt(0 , 0) y = -x

EXAMPLE 8 Find the equation of the line passing through A(0 , -5) & B(-3 , 1).

mAB =y2 – y1

x2 – x1

= 6-3

=1 – (-5)-3 – 0

= -2

Equation:

y = -2x - 5

Exercises from MIA book:

Page 8 Ex 4 Qu 3-12

Exercises from Heinemann book:

Page ? Ex ? Qu ??

EQUATION OF A STRAIGHT LINE 2

A is the point (a , b) and B is the point (x , y)

mAB =y2 – y1

x2 – x1

y – b

x – am =

m(x – a) = y - b

So we get:

y - b = m(x – a)

A

By

xO

Straight Line, Gradient m,passing through the

point (a , b)

To find the equation of a line now, we need only know the GRADIENT and ONE OTHER POINT –

we DO NOT require the y-intercept.

This version is much better than y = mx + c , because often it is difficult (and unnecessary) to write the

equation in this form.

EXAMPLE 9 Find the equation of the line passing through A(2 , 3) & B(6 , 6).

mAB =y2 – y1

x2 – x1

= 34

=6 – 36 – 2

y – b = m(x – a)

y – 3 = ¾(x – 2)

4y – 12 = 3(x - 2)

4y – 12 = 3x – 6

4y = 3x + 6

EXAMPLE 10

Find the equation of the straight line parallel to the line 2x + y + 3 = 0, and which passes through the point (3 , 4).

y – b = m(x – a)

y – 4 = -2(x – 3)

y – 4 = -2x + 6

y = -2x + 10

2x + y + 3 = 0

y = -2x – 3

So m =-2

Exercises from MIA book:

Page 11 Ex 6 Qu 2 - 6

Exercises from Heinemann book:

Page ? Ex ? Qu ??

MEDIANS AND ALTITUDES

MEDIAN A straight line from one vertex (corner) of a triangle to the

MIDPOINT of the opposite side.e.g. The line AP is a Median since

P is the midpoint of BC.

ALTITUDE A straight line from one vertex of a

triangle PERPENDICULAR to the opposite side.e.g. The line CQ is an Altitude as it is perpendicular to AB.

O

y

x

A

C

B

Q

P

EXAMPLE 11 Points P(4 , -4), Q(11 , 10) &R(-9 , 6) form a ∆. The median PS & altitude RT cross at K. Find K.P

R

Q

S

( )x1 + x2

2

y1 + y2

2S = ,

( )11 + (-9)

2S = , 10 + 6

2

S = (1 , 8)

mPS =y2 – y1

x2 – x1

= 12-3

=8 – (-4)1 – 4

= -4

y – b = m(x – a)

y – (-4) = -4(x – 4)

y + 4 = -4x + 16

y + 4x = 12

MEDIAN PS

P

R

Q

T

mPQ =y2 – y1

x2 – x1

= 147

=10 – (-4)11 – 4

= 2

ALTITUDE RT

Since perpendicular m1 x m2 = -1,

so mRT = -½

y – b = m(x – a)

y – 6 = -½(x – (-9))

2y - 12 = -x - 9

2y + x = 3

COORDINATES of K

y + 4x = 12 2y + x = 3

1

2

Sub x = 3 into y = -4x + 12 y = -4x3 + 12

1 x 2 2y + 8x = 24 3

3 2- 7x = 21

y = 0 y = -12 + 12

x = 3

P

R

Q

T

S

K

Therefore K is (3 , 0)

Exercises from MIA book:

Page 11 Ex 6 Qu 7-13

Exercises from Heinemann book:

Page ? Ex ? Qu ??

A straight line which BISECTS (cuts in half) another straight line & which is PERPENDICULAR to the line.

Perpendicular Bisectors

EXAMPLE 12 Find the equation of the perpendicular bisector of the line PQ, where P(7 , -12) & Q(-3 , 4).

( )x1 + x2

2

y1 + y2

2M = ,

( )7 + (-3)

2M = ,

-12 + 4

2

M = (2 , -4)

y – b = m(x – a)

y – (-4) = ⅝(x – 2)

8y + 32 = 5(x - 2)

8y + 32 = 5x – 10

8y - 5x = -42

Equation ofPerpendicular

Bisector:

mPQ =y2 – y1

x2 – x1

= 16-10

=4 – (-12)-3 – 7

= -85

Since perpendicularm1 x m2 = -1,

so mBIS = 5/8

Centroid

Useful Definitions

Orthocenter

Circumcenter

Point where the 3 Medians intersect!

Point where the 3 Altitudes intersect!

Point where the 3 Perpendicular Bisectors intersect! Centre of a circle joining all three vertex of the triangle!

Exercises from MIA book:

Page 14 Ex 7A/B All Qu.

Exercises from Heinemann book:

Page ? Ex ? Qu ??