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Linear Functions
● general form and graph shape
● using ​y=mx​+​c​ to convert between formulas, tables and graphs for linear functions
● finding the formula for a linear function from gradients and points on the line
● equation solution methods
● applications
● finding equations of lines on the Cartesian plane

Summary​ ​Learn​ ​Solve

 
 
Summary​  
A linear function is one which graphs as a straight line. It has general form 
​y​ = ​mx​ + ​c,​ where ​m​ is the gradient of the line and ​c​ is the ​y​-intercept.  
 
1. To find the formula for a line on a graph, given the gradient and a point on the 
line, start by assuming it is ​y​ = ​mx​ + ​c​, then substitute the gradient for ​m​, then 
substitute the ​x​ and ​y​ coordinates of the given point into the formula and solve 
the resulting equation to find ​c​. 
2. To find the formula for a line on a graph, given two points on the line, use the 
two points to find the gradient, then find ​c​ using one of the points as above. 
 
‘Find the equation of a line on the Cartesian plane’ is another way of saying 
‘Find the formula for a linear graph’. 
Because lines parallel to the ​y​ axis have neither gradient nor ​y​-intercept, they do not 
have formulae of the form ​y​ = ​mx​ + ​c​. Their formula is ​x​ = ​k​, where ​k​ is the ​x​-intercept 
of the line. 
 
 
 
 
 
 
 
 

 
Ms. Jona Cela Linear Functions Page 1
  
Learn​  
General form and graph shape  
Linear Functions ​are functions whose graphs are 
straight lines.   

We say the ​general form​ ​of a linear function is 

​y​ = ​c​ + ​mx.​   

● x​ is the independent variable ​which could be 

distance, number delivered etc.; 
● y​ is the dependent variable​ which could be fare, cost etc.;  
● c​ and ​m​ are numbers called ​parameters​. 

​The parameters are different in different linear functions. 

Stop for a while and make sure you understand the difference between variables and 
parameters – it’s a very important distinction. For example, if the variables are 
distance and cost, the distance determines the cost. 

Linear functions have two parameters, ​c​ and ​m.​   

● The value of ​c​ determines where the line 

crosses the vertical axis, the ​y​-intercept;  
● The value of ​m​ determines how steeply 
the graph rises from there, i.e. the gradient 
of the line.  
● The values of the parameters ​c​ and ​m​ can 
be fractional or negative.  
● If ​c​ is negative, then the line crosses the 
vertical axis below zero;  
● If ​m​ is negative, then the line slopes downwards rather than upwards. The 
function to the right is ​y​ = – 4 – ½​x​ . 

Although linear functions are probably easiest to understand in the form y = c 

+ mx (for example f = 3 + 2d), mathematicians more commonly write them as y 
= mx + c (e.g. f = 2d +3). 

 
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 Converting between formulae, tables and graphs 
The ​y​ = ​mx​ + ​c​ idea allows us to convert graphs and tables to formulae. 

● To convert a graph to a formula, just read off the ​y​-intercept: this gives you the 
value of ​c.​   
● Then determine the gradient: this gives you the value for ​m​.  
● Then substitute ​m​ and ​c​ into the general form ​y​ = ​mx​ + ​c​. 

For example, if the ​y​-intercept is 7 and the gradient is −2, then the formula is 
y ​ = −2​x ​ + 7. 
Practice 
Q1 Find the formulae for linear functions with: 
(a) gradient 2, ​y-​ intercept 5 (b) gradient 5, ​y​-intercept 1 
(c) gradient −3, ​y​-intercept 8 (d) gradient ½, ​y-​ intercept −3 
(e) gradient −0.6, ​y​-intercept −4 (f) gradient 1, ​y-​ intercept 3 
(g) gradient 0, ​y-​ intercept −1 (h) gradient 2, ​y-​ intercept 0 
(i) gradient −2, ​y​-intercept 0 (j) gradient 0, ​y​-intercept 0 

Exploration 1: 
Find the formula for the linear function in the case when the value of the independent 
variable (x) increases by 1 unit and starts at x=0. 

● To convert from a table to a formula, read off the value of ​y​ when ​x​ = 0: this will 
give the value of ​c.​   
● Then look at how much ​y​ increases for each unit increase in ​x​: this will give the 
value of ​m.​ (14-11=3; 17-14=3 etc) 
● Then substitute ​m​ and ​c​ into the general form ​y​ = ​mx​ + ​c​. 
x 0 1 2 3 4 5
y 11 14 17 20 23 26
In the table above ​c​ = 11 and ​m​ = 3, so the formula is ​y​ = 3​x​ + 11. 

Exploration 2: 
Find the formula for the linear function in the case when the value of the independent 
variable (x) increases by 1 unit but does not start at x=0. 

● In the second table (above), we can see that ​m​ = −2 (8-6=2; 6-4=2 etc), but we 
have to move backwards to find ​c​.  
● Imagine the pattern continuing back to ​x​ = 0; the value of ​y​ would be 14. 
● So ​c​ = 14. So the formula is ​y​ = −2​x​ + 14.  

x 3 4 5 6 7 8
y 8 6 4 2 0 −2

 
Ms. Jona Cela Linear Functions Page 3
 Exploration 3: 
Find the formula for the linear function in the case when the value of the independent 
variable (x) increases ​not anymore​ by 1 unit and does not start at x=0. 

● In this last table, the ​x​ values go up in 4s.  

● First we isolate the first two points given in this table (6, 44) and (10, 56).  
● We subtract the y-s, 56-44=12, then we subtract the values of the x-s, 10-6= 4.  
● Now we divide the difference of the y-s with the difference of the x-s: ​12 ÷ 4 = 3, 
so ​m​ = 3. 
● We have to extrapolate back from the first ​y-​ value. This would be 6 steps of 3 
taking us back 18 to 26. So ​c​ = 26. And the formula is ​y​ = 3​x​ + 26. 
x 6 10 14 18 22 26
y 44 56 68 80 92 104
 
Practice 
Q2 Find the formulae for these linear functions.  

(a)  x 0 1 2 3 4 5
  y 5 8 11 14 17 20
 
(b)  d 3 4 5 6 7 8
  f 10 12 14 16 18 20
 
(c)  t 0 2 4 6 8 10
  r 12 6 0 -6 -12 -18
 
(d)  x 8 12 16 20 24 28
  y 5 7 9 11 13 15
 
(e)  x -2 -1 0 1 2 3
  y -13 -11 -9 -7 -5 -3
 
(f)  x 2.5 3 3.5 4 4.5 5
  y 28 32 36 40 44 48
 

 
 
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 Plotting the graph of a linear function 

The general form ​y​ = ​mx​ + ​c​ ​can also be used as a quick way to plot the graph of a 
linear function from its formula without having to draw up a table. 
 
Way 1 

● To plot the graph of ​y​ = 3​x​ – 4, place a dot on the ​y-​ axis at -4.  
● From there step 1 unit to the right and 3 units up and place another dot.  
● Then, for accuracy, go another unit right and another 3 units up and place 
another dot. Then join the dots.  

Way 2 

● To plot the graph of ​y​ = 3​x​ – 4, place a dot on the ​y-​ axis at -4.  
● Give a value of your choice to x and substitute it in the equation to find the 
respective y value. ( x=3, y=3*3-4=9-4=5,  
● Then plot beside (0,4) the second point that we calculated (3,5). Join the dots 
with a continuous line.  
Practice 
Q3 Use the ​y​ = ​mx​ + ​c​ idea to quickly plot graphs of the following functions. 
(a) ​y​ = 2​x​ + 5 (b) ​y​ = 2​x​ − 3 
(c) ​y​ = ​x​ + 1 (d) ​y​ = −½​x​ + 2 
(e) ​y​ = 2​x (f) ​y​ = 6

 
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 Finding formulae from gradients and points on the line 
Given the gradient and a point on the line 
 

 
● Suppose a function has a gradient of 2 and it passes through the point (5, 3). 
● m​ = 2, so the formula is ​y = 2 ​ ​x ​+ ​c​. 
● Now when ​x​ = 5, ​y​ = 3 according to the given point (5,3).   
● We can substitute these into ​y =​ 2​x​ + c to get 3 = 2 ​×​ 5 +​ c.​    
● We can then solve this equation to get c = –7.   
● So the formula is ​y = 2​ ​x​ – 7. 
Practice 
Q4 Find the formulae for the following linear functions: 
(a) gradient = 2,  passes through (4, 1) 
(b) gradient = –1,  passes through (–6, 2) 
(c) gradient = ¼,  passes through (–8, –3) 
(d) gradient = 0.8,  passes through (5, 0) 
(e) gradient = 0,  passes through (3, 3) 

 
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 Given two points on the line 
Suppose we want the formula for the line that passes through the two points (–2, 7) 
and (4, 1). 

● First we find the rise and run between the two points ( we have to divide the 
difference between the y-s ​(we call this rise)​ with the difference between the 
x-s ​(we call this run)​). 
● Then we calculate the gradient from these using the riserun method. Then we 
know the gradient and a point on the line (we can choose either of the given 
points). So we can find the formula using the method above. 

● In the graph above, the rise is –6, the run is 6, so the gradient is –1. So ​m​ = –1. 
● So the formula is ​y = – x + c.​    
● Then we can use the point (4, 1) to find ​c​.   
● We sub ​y ​= 1, ​x​ = 4 into ​y = – x + c​. This gives us 1 = –4 + c. Hence c = 5. 
● So the formula is ​y = – x +​ 5 

 
Practice 
Q5 Find the formulae for the following linear functions: 
(a) passes through (1, 2) and (5, 10) 
(b) passes through (–6, 2) and (2, –2) 
(c) passes through (–8, –3) and (5, 10) 
(d) passes through (5, 0) and (–5, 4) 
(e) passes through (3, 3) and (–5, –2) 

 
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 Solving Equations 
If we substitute for the dependent variable in a linear function, we get an equation 
something like 3​a​ + 5 = 23. You already know how to solve such equations by doing 
the same thing to both sides.  
3​a​ + 5 = 23 
−5 −5 

3​a​ = 23 
​÷3 ÷3 

​a​ = 7.67 

You have done it enough that you shouldn’t need any more practice here. 

Vertical Lines 
One exception to this is the equation of a vertical line (parallel to the ​y-​ axis). Such 
lines have an infinite gradient and no ​y​-intercept, and so can’t be expressed in the 
form ​y​ = ​mx​ + ​c​. They are not linear functions. They don’t pass the vertical line test for 
functions. 

The equation of a vertical line is just ​x​ = ​k,​ where ​k​ is the ​x​-intercept. Line (d) on the 
plane above has the equation ​x​ = 9.5. It is all the points on the plane where the 
x-​ coordinate is 9.5. 

Parallel and Perpendicular lines 

Parallel lines on the Cartesian plane have the same gradient​ and therefore the 
same value for the parameter ​m​. So ​y​ = 3​x​ + 5 and ​y​ = 3​x​ – 4 are parallel. 

If lines are perpendicular, then the gradient of one is the negative of the 
reciprocal of the other.​ So if one has a gradient of 4, the other has a gradient of –¼; 
if one has a gradient of −​2​/​3​, then the other has a gradient of 3​​ /​2​. To get the reciprocal of 
any number, express it as a proper or improper fraction and turn it upside down.  

 
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 Practice 
Q6 (a) Find the equation of a line on the Cartesian plane which is vertical  
(parallel to the ​y-​ axis), but 4 units to the right of it. 
(b) Find the equation of a line on the Cartesian plane which is parallel to  
the ​y-​ axis), but 2 units to the left of it. 

Q7 (a) Give the gradient of a line which is parallel to a line with gradient −0.7. 
(b) Give the gradient of a line which is perpendicular to a line with gradient  
2. 
(c) Give the gradient of a line which is perpendicular to a line with gradient  
−4. 
(d) Give the gradient of a line which is perpendicular to a line with gradient 
​ /​5​. 
2​

(e) Give the gradient of a line which is perpendicular to a line with gradient  
−1½. 

Q8 (a) Find the equation of the line parallel to ​y​ = 2​x​ + 3 but passing through the 
point (4, 1). 
(b) Find the equation of the line parallel to ​y​ = −​x​ + 3 but passing through the  
point (6, 5).  
(c) Find the equation of the line perpendicular to ​y​ = 2​x​ − 1 but passing  
through the point (4, −1).  
(d) Find the equation of the line perpendicular to ​y​ = ½​x​ + 29 but passing  
through the point (−2, −3). 

 
 
Solve​  
Q51 A pizza company charges a set price per pizza plus a set price per order for the 
delivery. Getting 3 pizzas costs $23; getting 10 pizzas costs $58. How much 
would it cost for 34 pizzas? 

Q52 By plotting the lines ​y​ = 2, ​y​ = 0.5​x​ + 3 and ​y​ = −​x​ + 9, find the area of the 
triangle between them. 

Q53 By finding the equations of the 3 lines connecting the points (4, 3), (9, 5) and 
(7, 10), show that they form a right-angle triangle and show that the area of the 
triangle is 14.5 units. 

Q54 How far (in a straight line) is it from the point (−3, −1) to the intersection of the 
line ​y​ = ​x​ + 4 and the line ​x​ = 5? 

Q55 Find the equation of the line with a positive gradient that crosses the ​x​-axis at 
−4 and makes an angle of 20° with the ​x​-axis. 
 
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