Linear Relationships
4.1 Straight Line Graph
4.1.1
Introduction
The straight line is one of the most common mathematical representations used in science. A ‘straight line’, or linear, relationship occurs when the change in one variable (e.g.y) is proportional to the change in another variable (e.g.x), and is commonly represented by the
straight line equation:y= mx + c.
The slope,m, of the line gives the rate of change of y with respect to x. The point at which
the line crosses they-axis is called the intercept, c – see Figure 4.2 below.
y-axis x-axis y1 y2 y2−y1 x2 x1 x2−x1 (x2, y2) (x1, y1) y = mx + c c q 0
Figure 4.2 Straight line graph.
4.1.2
Plotting the graph
In an experiment, the independent variable is the one whose values are chosen, and the
dependent variable is the one whose values are measured . For example, the pH (dependent
variable) of the soil might be measured for known amounts (independent variable) of added lime.
Anx –y graph is normally plotted with x as the independent variable on the horizontal axis
(abscissa) andy as the dependent variable on the vertical axis (ordinate).
Q4.1
In an experiment to measure plant growth as a function of light exposure, different plants are exposed to different levels of light exposure, L, and the resultant
growth,G, is recorded.
When the results are plotted, which of the following statements are True? (i) G should be plotted against L (i.e. L on the x-axis).
(ii) L should be plotted against G (i.e. G on the x-axis).
(iii) It does not matter which way round the results are plotted.
4.1.3
Straight line equation
A point is located on an x –y graph by using its co-ordinates (x, y). Note that the x-value
is placed first.
Only one straight line can be drawn through any two particular points, as illustrated in Figure 4.2.
In this book we will describe the straight line by using the common equation:
y = mx + c [4.5]
wherem and c are constant values that define the particular line:
• Slope of the line is given by the coefficient of x: m.
• Intercept on the y-axis when x= 0 is given by the constant: c. [4.6]
Example 4.1
Calculate the slope and intercept of the straight line described by the equation: 2y = 4x + 3
The first step is to rearrange the equation into the form y= mx + c:
Divide both sides of the equation by ‘2’: y= (4/2)x + (3/2)
giving: y= 2x + 1.5
Comparing with the standard equation, [4.5]: y= mx + c
Q4.2
Start by plotting the line y= 2x + 3 on a graph over the range −2 < x < +2:
(i) For each of the values ofx in the table below, calculate the value of y given
by the equationy= 2x + 3; this gives the co-ordinates (x, y) of points that
will be on this straight line.
x −2 −1 0 +1 +2
y
(ii) Plot the co-ordinates from (i) on anx –y graph, and connect with a line.
(iii) Where does the graph intercept they-axis?
Do the following points lie on the line?
(iv) (1.5, 5.5) Yes/No
(v) (0.5, 4) Yes/No
(vi) (−0.5, −1.0) Yes/No
Without carrying out any new calculations, sketch the following lines on the same graph as (ii):
(vii) y= −2x + 3
(viii) y= 2x + 1
(ix) y= x + 3
Q4.3
The length, L (m), of a simple metal pendulum as a function of the ambient
temperature,T (◦C), is given by the equation:
L= (1 + α × T ) × 1.210
whereα is the coefficient of linear expansion of the pendulum material, and 1.210
is the length of the pendulum when T = 0◦C. Assumeα= 0.000 019◦C−1. (i) Multiply out the bracket to obtain a straight line equation of the form
y = mx + c.
(ii) Calculate the slope,m, and intercept, c, of the equation in (i).
Q4.4
A straight line is given by the equation: 3x+ 4y + 2 = 0
Rearrange the equation to makey the subject of the equation. Hence, calculate the
slope and intercept of the line.
If a particular point with co-ordinates (x, y) lies on the line, then the values of x and y will
make the equation balance, i.e. the value ofy will equal the value of mx+ c and the equation
will be TRUE.
If the point (x, y) does not lie on the line, then the equation will not balance, and the value
ofy will not equal the value of mx+ c. Example 4.2
The point (3,10) lies on the straight liney = 2x + 4:
Check by replacingx with ‘3’: 2x+ 4 ⇒ 2 × 3 + 4 ⇒ 6 + 4 ⇒ 10
which balances withy= 10 – the equation is TRUE.
The point (4,14) does not lie on the straight line y= 2x + 4:
Check by replacingx with ‘4’: 2x+ 4 ⇒ 2 × 4 + 4 ⇒ 8 + 4 ⇒ 12
which does not balance with y= 14.
It is often necessary to calculate a value forx, given a value of y on a known straight line.
We need then to rearrange equation [4.5] to makex the subject of the equation:
x =(y− c)
m [4.7]
Q4.5
A straight line (y = mx + c) has a slope of +4 and an intercept of −3. Calculate
the value ofx when y = 4
Q4.6
The cooking time for a joint of meat is written as 40 minutes per kilogram plus 20 minutes.
(i) Express this as an equation relating time,T (in minutes), and the mass, W (kg).
(ii) If there is only 2 hours available, what is the heaviest joint that could be cooked?
Example 4.3
A car is travelling at a constant speed of 30 m s−1, along a straight road away from a junction. I start a stopwatch with t = 0 when the car is 60 m away from the
junction.
(i) Write down an equation that will then relate the distance,z, of the car from the
junction and the time,t, in seconds on my stopwatch.
(ii) Calculate the distance of the car from the junction whent= 3 s.
(iii) What will be the time when the car is 210 m from the junction?
The analysis is performed in the following text .
The line and calculations for Example 4.3 can be represented on a graph ofz against t:
0 50 100 150 200 250 300 350 0 1 2 3 4 5 6 7 8 9 10 Time, t Distance, z
Figure 4.3 Graph for Example 4.3.
(i) The rate of change of distance,z, with time, t, is 30, and this will be a slope, m= 30 in
the equation – see Figure 4.3.
The intercept ,c, is given by the value of z when t is zero. This is given in the question
as 60 m. Hencec= 60.
The equation is therefore:
z= 30 × t + 60
(ii) Whent = 3, we can use the above equation to calculate:
z= 30 × 3 + 60 ⇒ 90 + 60 ⇒ 150 m
(iii) To find a value fort it is necessary to rearrange the above equation into the form given
by [4.7]:
t = (z− 60)
30
Substituting values gives:t= (210 − 60)/30 ⇒ 150/30 ⇒ 5 s.
Q4.7
A car is travelling at a constant speed of 20 m s−1 along a straight road towards a junction. It is at a distancez= 140 m away from the junction when I start my
stopwatch at timet = 0 s.
(i) Which of the following equations will now describe the motion of the car? (a) z= 20t + 140
(b) z= 20t − 140
(c) z= −20t + 140
(d) z= −20t − 140
(ii) What will be the time when the car passes the junction? (iii) What will be the position of the car whent = 20 s?
4.1.4
Calculating slope and intercept
The slope, m, of the straight line (Figure 4.2) that passes through points (x1, y1) and (x2, y2) is given by the equation:
m=(y2− y1) (x2− x1)
or m= (y1− y2) (x1− x2)
[4.8]
The intercept, c, of the straight line is the value ofy at the point where the line passes through
they-axis (Figure 4.2), i.e. the value of y when x= 0.
To calculate the equation of a straight line that passes through the two points (x1, y1) and (x2, y2):
[4.5] – this gives an equation where the intercept,c, is the only unknown value. Rearrange
the equation to makec the subject:
c= y1− m × x1 or c= y2− m × x2
and calculate the value of the intercept,c.
Example 4.4
Calculate the equation of the line that passes through the points (−2, 3) and (2, 1). Calculating the slope:
m=(y2− y1) (x2− x1) ⇒ (1− 3) (2− (−2)) ⇒ −2 (2+ 2) ⇒ −2 4 ⇒ −0.5
Substitutem into the equation y2 = mx2+ c, with the co-ordinates x2= 2 and y2 = 1: 1= (−0.5) × 2 + c ⇒ −1 + c
Rearranging gives the intercept:
c= 1 + 1 = 2
The equation of the straight line is therefore:
y = −0.5x + 2
Q4.8
Calculate the slopes of the straight lines which pass through each of the following pairs of points:
(i) (1, 1) and (2, 3) (iv) (2.0, 3.0) and (1.0, 3.5)
(ii) (−1, 1) and (2, 3) (v) (0, 1) and (1, 0)
(iii) (1, 1) and (2,−3) (vi) (2.0, 3.0) and (1.0,−3.5)
In some cases where it is necessary to calculate the coefficients of a straight line, the slope,
m, may already be known. It is then possible to go directly to step 2 in the calculation process
Q4.9
Derive the equations of the straight line that has: (i) a slope of 2.0 and passes through the point (0, 3) (ii) a slope of 0.5 and passes through the point (2, 3) (iii) a slope of −0.5 and passes through the point (2, 3)
Q4.10
IfC represents the temperature in degrees Celsius and F represents the temperature
in degrees Fahrenheit, water boils at a temperature given byC= 100 and F = 212
and water freezes at a temperature given byC= 0 and F = 32. Derive a straight
line equation which will give F as a function of C.
Hence calculate the value of C when F = 0
(hint: F and C become the y- and x-axes of a graph).
4.1.5
Intersecting lines
If two straight linesy= mAx+ cAandy = mBx+ cBmeet at a point (x0, y0), then the values ofx0 andy0 will make both equations true simultaneously :
y0= mAx0+ cA
y0= mBx0+ cB
Any unknown variables can (usually) be found by the methods of simultaneous equations.
Example 4.5
Calculate the co-ordinates of the crossing point for the two straight lines:
y = −0.5x + 2 y = x − 1
Using simultaneous equations (3.5.3) to ‘solve’ the above equations, we find that the values x= 2 and y = 1 will make both equations true simultaneously.
Substituting these values back into the above equations makes both equations TRUE. The point (2, 1) therefore lies on both lines. This is the point where the lines cross.
4.1.6
Parallel and perpendicular lines
Parallel lines have the same slope:
mA= mB [4.9]
(but different intercepts,cA= cB).
Perpendicular lines have slopesmA andmB given by the relationships:
mA= − 1 mB andmB= − 1 mA [4.10] Example 4.6
Calculate the equation of the line which is perpendicular to the liney = −0.5x + 2 and
passes through the point (−2, 3). The slope of the first line,mA= −0.5
Thus the slope of the perpendicular line,mB= − 1 mA ⇒ − 1 −0.5 ⇒ 2
Substitute this value ofm into the equation y= mBx+ cB, with the co-ordinates (−2, 3), and calculate the value of the intercept,cB:
3= 2 × (−2) + cB⇒ −4 + cB Hence cB= 7, and the equation of the line becomes: y = 2x + 7
Q4.11
A second line is parallel to the line y= 3x + 2, and passes through the point
Q4.12
A second line is perpendicular to the line y = 3x + 2, and passes through the
point (1,1). What is the equation of this second line?
Example 4.7
Write down the equation of the line that is parallel to thex-axis (horizontal) and passes
through they-axis at y= 2.5.
A line that is parallel to the x-axis has slope m= 0.
If the line passes through they-axis at y= 2.5, then c = 2.5.
Hence the equation of line is:
y = 2.5
Whatever the value of x, the horizontal line gives the same value, y= 2.5.
Q4.13
Write down the equation of the line that is parallel to the y-axis (vertical) and
passes through thex-axis at x= −1.5.
4.1.7
Interpolation and extrapolation
These are defined as follows:
Interpolation – finding the co-ordinates of a point on the line between existing points. Extrapolation – finding the co-ordinates of a point on the line outside existing points.
In both interpolation and extrapolation, it is necessary to calculate first the equation of the line that passes through the two points, and then find the value ofx (or y) equivalent to a new
value ofy (or x).
Q4.14
A straight line passes through the points (2, 3) and (4, 4). Calculate the values of: (i) y when x = 3.8 (iii) y where the line crosses the y-axis
Q4.15
Initially, at timet = 0, the height, h, of a plant is 15 cm, and it then grows linearly
over a 20-day period, reaching a height of 20 cm. Derive a straight line equation which will give the height,h, as a function of the time, t, in days.
(i) What is the height whent = 5 days?
(ii) Estimate the height of the tree 8 days after the end of the 20-day period, assuming that it continues to grow at the same rate.
4.1.8
Angle of slope
The angle,θ , between the line and the x-axis (Figure 4.2) is given by the equation:
tan(θ )=(y2− y1) (x2− x1)
= m [4.11]
The angle,θ , can then be calculated using the inverse tangent function (2.4.7):
θ= tan−1(m) or arctan(m) [4.12]
Q4.16
A map shows the height contours on the side of a hill of fairly uniform slope. A point on the 250 m contour line is seen to be 400 m horizontally from a point on the 200 m contour line.
(i) Calculate the average slope,m, between the two points.
(ii) Calculate the average slope angle of the ground with respect to the horizontal.