THE STRAIGHT LINE
Horizontal & Vertical Lines
Horizontal lines give an equation of the form y = a because all of the
y-coordinates are the same.
Vertical Lines give an equation of the form x = a because all of the
x-coordinates are the same.
Ex1
Find the horizontal and vertical lines through the point (1,3).
The horizontal line has equation y = 3
The vertical line has equation x = 1
Gradient
In general maths:
Ex1
Ex2
A car travels up a hill with a gradient of
travels a horizontal distance of 120m?
Vertical Height
Horizontal Distance
1
x
=
5 120
1
120 × = x
5
24 = x
Gradient =
∴ It travels 24metres up
1
how far will it go up when it
5
The Gradient Formula
In Credit maths:
Gradient (m) =
y 2 − y1
x 2 − x1
Where (x1 , y1 ) and (x2 , y2 ) are points on the line.
Ex1
Find the gradient of the line PQ.
x1 y 1
x2 y2
P (3 , 1) Q(6 , 2)
mPQ =
2 −1
6−3
1
=
3
=
mPQ
y 2 − y1
x 2 − x1
Ex2
Find the gradient of the line RS.
x 1 y1
x2 y2
R ( −3 , 1) S(1 , -7)
mRS =
y2 − y1
x 2 − x1
−7 − 1
1 − ( −3)
−8
=
4
= −2
=
mRS
Remember: mRS = mSR
Drawing Lines
In order to draw a line we need to find any two points on it. This can be
done by picking any two values of x and fitting them in to get the y
coordinates.
Ex1.
Sketch the lines :
a) y = 2x – 1
b) y = x - 2
c) y = -2x + 2
(a) y = 2x - 1
When x = 1
y = 2(1) − 1 = 1
(1,1)
When x = 2
y = 2(2) − 1 = 3
(2,3)
(b) y = x − 2
When x = 1
y = 1 − 2 = −1
(1,-1)
When x = 3
y = 3−2 =1
(3,1)
(c) y = −2x + 2
When x = 0
y = −2(0) + 2 = 2
(0,2)
When x = 2
y = −2(2) + 2 = −2
(2,-2)
The Equation of a Line
The equation of the line through the point (0, c) with gradient m is :
y = mx + c
The point (0, c) is where the graph crosses the y axis and is called the
y-intercept.
Ex1
What is the equation of the line with gradient 4, passing through the point
(0, 3) ?
Ex2
Find the gradient and the y-intercept of the line with equation 3x + 4y = 5
Finding the Equation Given Any Two Points
Ex1
Find the equation of the line through the points (2, 5) and (-1, -4)
Modelling Situations
This is similar to the last two exercises except that the x and y axes are
replaced with different letters.
m =8
c = 30
y = mx + c
y = 8x + 30
C = 8T + 30
This is the equation of the line and also a formula for the cost given the
time from the graph.
Line of Best Fit (Scatter Diagrams)
In chapter 6 (book 3) we used the following example to draw a scatter
diagram.
Ex –
The shoe sizes and heights of 20 pupils are:
Pupil
1
2
3
4
5
6
7
8
9
10
Shoe
Size
7.5
8.5
7
8
10
6.5
6
8.5
9.5
8
Height
135
143
131
145
160
135
120
147
152
140
Pupil
11
12
13
14
15
16
17
18
19
20
Shoe
Size
9
7
9
7.5
6
8
8.5
9.5
10
6.5
Height
150
135
152
143
129
137
150
155
155
126
This appears to show a positive correlation between shoe size and height.
It is possible to draw a
line of best fit which
passes as close to as many
points as possible.
Once this is drawn you
use the line to answer any
other questions.
In addition you can also
now find the equation of
the line.
How to find the equation of the line of best fit:
First you need the gradient –
Start by choosing two well separated points.
1
So y = 8 x + c
3
Now fit in one of the points. E.g. (6,125)
1
125 = 8 × 6 + c
3
125 = 50 + c
c = 125 − 50
c = 75
1
Now the equation of the line becomes: y = 8 x + 75
3