Straight line

Higher Unit 1
www.mathsrevision.com

Higher

Outcome 1

Distance Formula
The Midpoint Formula
Gradients
Collinearity
Gradients of Perpendicular Lines
The Equation of a Straight Line
Median, Altitude & Perpendicular Bisector
Exam Type Questions

Starter Questions
Outcome 1


Higher

2.

1.

Calculate the length of the length AC.
A
6m
Calculate the coordinates
that are halfway between.

(a) ( 1, 2) and ( 5, 10)

(b)

B

8m

C

( -4, -10) and ( -2,-6)

Distance Formula
Length of a straight line
Outcome 1


Higher

AB =AC +BC
2

2

2

y
B(x2,y2)

This is just

y2 – y1

A(x1,y1)

Pythagoras’ Theorem

x2 – x1
O

C

x

Distance Formula

Higher

Outcome 1

The length (distance ) of ANY line
can be given by the formula :
ABdis tan ce = (y2 − y1 ) + (x2 − x1 )
Just
Pythagoras
Theorem in
disguise

Finding Mid-Point of a line
Outcome 1


Higher

The mid-point between
2 points is given by
Simply add both x
coordinates together
and divide by 2.
Then do the same with
the y coordinates.

y
y2

A(x1,y1)

M

y1
O

 x1 + x 2 y1 + y 2 
M =
,
,÷
2
 2


B(x2,y2)

x1

x2

x

Straight line Facts

Higher

Outcome 1

y = mx + c
y2 - y1
Gradient =
x2 - x1

Y – axis
Intercept

Another version of the straight line general formula is:
ax + by + c = 0

Outcome 1

Higher


Sloping left to right up has +ve gradient
m>0

Sloping left to right down has -ve gradient

m<0

Horizontal line has zero gradient.

m=0

y=c

Vertical line has undefined gradient.
x=a

Feb 2, 2014

7

Outcome 1


Higher

Lines with the same gradient
m>0

means lines are Parallel
The gradient of a line is ALWAYS
equal to the tangent of the angle

θ

made with the line and the positive x-axis

m = tan θ

Feb 2, 2014

8

Collinearity
Outcome 1

Higher


Points are said to be collinear
if they lie on the same straight.

y

C

The coordinates A,B C are
collinear since they lie on
the same straight line.

B

D,E,F are not collinear they
do not lie on the same
straight line.

A

E

F

D
O

x1

x2

x

Gradient of perpendicular lines
Outcome 1

Higher

→ B (-b, a)


When rotated through 90º about the origin A (a, b)

mOB

a-0
a
=
=-b - 0
b

B(-b,a)

y -a
-b

-b

A(a,b)

O

mOA ×mOB =

a

mOA

b-0 b
=
=
a -0 a

x

b a
ab
×
== -1
a -b
ab

If 2 lines with gradients m1 and m2 are perpendicular then m1 × m2 = -1
Conversely:
If m1 × m2 = -1 then the two lines with gradients m 1 and m2 are perpendicular.

The Equation of the Straight Line

y – b = m (x - a)
Outcome 1

Higher


The equation of any line can be found if we know
the gradient and one point on the line.
y

P (x, y)

y
m

b
O

A (a, b)

a

x–a

y-b
m=
(x – a)

y -- b
b
x

x

Gradient,

y–b=m(x–a)

m

Point

(a, b)

Point on the line ( a, b )

Typical Exam Questions
Outcome 1


Higher

Find the equation of the line which passes through the point (-1, 3)
and is perpendicular to the line with equation 4 x + y − 1 = 0
Find gradient of given line:

4 x + y − 1 = 0 ⇒ y = −4 x + 1 ⇒ m = −4

Find gradient of perpendicular:
Find equation:

m=

1
(using formula m × m = −1)
1
2
4

y – b = m(x – a)
y – 3 = ¼ (x –(-1))
4y – 12 = x + 1
4 y − x − 13 = 0

Outcome 1


Higher

Find the equation of the straight line which is parallel to the line

2 x + 3with 5
y = equation
and which passes through the point (2, –1).

Find gradient of given line:

2
3

3 y = −2 x + 5 ⇒ y = − x + 5 ⇒ m = −

Gradient of parallel line is same:
Find equation:

M = -2/3

y – b = m(x – a)
y – (-1) = -2/3 (x – 2)
3y + 3 = -2x + 4
3 y + 2x =1

2
3

Outcome 1

Higher


Median means a line from vertex
to midpoint of the base.

A

A
B

D

C

D
Altitude means a perpendicular line

B

C

from a vertex to the base.


Feb 2, 2014

14

Outcome 1


Higher

Perpendicular bisector means a line from the vertex
that cuts the base in half and at right angles.

A

B

D

C
Feb 2, 2014

15

Exam Type Questions
Outcome 1


Higher

Find the size of the angle a° that the line
joining the points A(0, -1) and B(3√3, 2)
makes with the positive direction of the x-axis.

2 − (− 1)
3
=
=
Find gradient of the line: m =
3 3−0
3 3
m = tan θ

tan θ =

Use table of exact values

1
3

θ = tan

−1

1
θ = 30°
3

1
3

Exam Type Questions
Outcome 1

Higher


A and B are the points (–3, –1) and (5, 5).
Find the equation of
a) the line AB.
b) the perpendicular bisector of AB
3
Find equation of AB 4 y = 3 x + 5
Find gradient of the AB: m =

4

Find mid-point of AB

(1,2)

4
Gradient of AB (perp): m = −
3

Use gradient and mid-point to obtain perpendicular bisector AB
4x + 3y = 10

Outcome 1


Higher

π
The line AB makes an angle of
radians with
3
the y-axis, as shown in the diagram.
Find the exact value of the gradient of AB.

π π π
− =
Find angle between AB and x-axis:
2 3 6
π
m = tan θ
m = tan
6
Use table of exact values

1
m=
3

(x and y axes are perpendicular.)

Higher

Outcome 1


A triangle ABC has vertices A(4, 3), B(6, 1)
and C(–2, –3) as shown in the diagram.
Find the equation of AM, the median from A.

 x2 - x1 y2 − y1 
,
Find mid-point of BC: (2, − 1) Using M 
÷
2
2 

Find gradient of median AM m = 2 Using m =

y2 - y1
x2 - x1

Find equation of median AM y = 2 x − 5 Using y - b = m( x - a )


Higher

Outcome 1

P(–4, 5), Q(–2, –2) and R(4, 1) are the vertices
of triangle PQR as shown in the diagram.
Find the equation of PS, the altitude from P.
Find gradient of QR: m =

1
y -y
Using m = 2 1
2
x2 - x1

Find gradient of PS (perpendicular to QR)

m = − 2 (m1 × m2 = − 1)

Find equation of altitude PS

y + 2x + 3 = 0

Using y − b = m( x − a )


Higher

Typical Exam
Questions

72o

Outcome 1

The lines y = 2 x + 4

63

and x + y = 13

o

45

o

135o

make angles of a° and b° with the positive direction of the xaxis, as shown in the diagram.
a) Find the values of a and b
b) Hence find the acute angle between the two given lines.

y = 2x + 4

m=2

Find a°

tan a° = 2 → a = 63°

x + y = 13

m = −1

Find b°

tan b° = − 1 → b = 135°

Find supplement of b

= 180 − 135 = 45°

Use angle sum triangle = 180° angle between two lines

72°

Higher

Exam Type
Questions Outcome 1

p
q


Triangle ABC has vertices
A(–1, 6), B(–3, –2) and C(5, 2)
Find:

a) the equation of the line p, the median from C of triangle ABC.
b) the equation of the line q, the perpendicular bisector of BC.
c) the co-ordinates of the point of intersection of the lines p and q.
Find mid-point of AB

(-2, 2)

Find equation of p

y=2

Find mid-point of BC

(1, 0)

Find gradient of q

m = −2

Find gradient of p

m=0

1
Find gradient of BC
2
y = −2 x + 2
Find equation of q

Solve p and q simultaneously for intersection (0, 2)

m=


Higher

Exam Type
Questions

l2
Outcome 1

l1

Triangle ABC has vertices A(2, 2), B(12, 2) and C(8, 6).
a) Write down the equation of l1, the perpendicular bisector of AB
b) Find the equation of l2, the perpendicular bisector of AC.
c) Find the point of intersection of lines l1 and l2.
Mid-point AB

( 7, 2 )

Find mid-point AC
Gradient AC perp.
Point of intersection

Perpendicular bisector AB

x=7

2
(5, 4) Find gradient of AC m =
3
3
m=−
Equation of perp. bisector AC 2 y + 3 x = 23
2

(7, 1)


Higher

Exam Type
Questions

Outcome 1

A triangle ABC has vertices A(–4, 1), B(12,3) and C(7, –7).
a) Find the equation of the median CM.
b) Find the equation of the altitude AD.
c) Find the co-ordinates of the point of intersection of CM and AD
Mid-point AB

( 4, 2 )

Equation of median CM
Gradient BC
Equation of AD

m=2

Gradient CM (median)

m = −3

y + 3x = 14
Gradient of perpendicular AD

2y + x + 2 = 0

Solve simultaneously for point of intersection (6,

-4)

1
m=−
2

Higher

Exam Type
Questions

M

Outcome 1


A triangle ABC has vertices A(–3, –3), B(–1, 1) and C(7,–3).
a) Show that the triangle ABC is right angled at B.
b) The medians AD and BE intersect at M.
i) Find the equations of AD and BE. ii) Find find the co-ordinates of M.
Gradient AB

m=2

Product of gradients

Gradient BC

2 × −

1
→ 1
2

1
m=−
2

Hence AB is perpendicular to BC, so B = 90°

1
Equation AD 3 y − x + 6 = 0
3
4
2, − 3 ) Gradient of median BE m = − Equation AD 3 y + 4 x + 1 = 0
Mid-point AC (
3
5

Solve simultaneously for M, point of intersection  1, − ÷
3

Mid-point BC ( 3, − 1)

Gradient of median AD m =

Straight line

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