LINEAR FUNCTIONS
LINEAR FUNCTIONS
A linear function is of the type y = mx + c. Examples of linear functions are:
y = 4x,
y = 5 – 3x,
𝑥
𝑥
𝑦 = 2,
4
𝑦
+5=2
For pairs of values of x and y as shown below, we can say it is linear if the first
differences are same.
x
y
2
7
First difference
3
10
4
13
3
3
5
16
3
What is the rule?
The rule is y = 3x + 1.
State which of the following is linear functions.
Function
Linear or Not
y = 2x – 1
𝑦=
YES
3
𝑥
4
y = 7 + 4x
y2 = 2x
5
y=𝑥
𝑦 = 𝑥2 + 4
𝑥
= 𝑦−4
2
Page 1
6
19
3
LINEAR FUNCTIONS
EXERCISE 3.1
Investigate which of the following are linear functions. If yes, find the rule without a
calculator.
(a)
x
y
1
3
2
5
3
7
4
9
5
11
6
13
x
y
-1
7
0
10
1
13
2
16
3
19
4
22
x
y
3
10
5
15
7
20
9
31
11
40
0
10
1
8
2
6
3
4
(b)
(c)
(d)
x
y
Page 2
LINEAR FUNCTIONS
Tabulate
Use your calculator to complete a table of values.
1. Complete the following tables using your calculator.
x
y = 2x
-2
-4
-1
0
0
1
2
5
10
x
y = 3x
-3
-9
-2
0
0
1
4
12
10
x
y = 7 - 2x
0
7
1
2
3
1
4
5
x
y = 4x + 1
2
3
13
5
8
33
10
20
x
y = 8 - 3x
0
1
2
2
3
4
5
-7
2. Complete the following tables using your CAS calculator.
x
y = 5x - 2
2
8
3
5
8
38
10
20
x
y = 3x - 5
-2
-11
-1
0
1
3
5
x
y=x+5
-4
1
-3
-2
3
0
5
8
Page 3
LINEAR FUNCTIONS
FINDING GRADIENT AND Y-INTERCEPT
Express each of the following in the form y = mx + c and hence state the gradient (m) and
the y-intercept (c). Remember to obtain y by itself on the left-hand side, then pick up the
value of m and c.
Equation
Gradient (m)
y = 2x + 5
y = 3x + 7
y=x-3
y = 0.5x + 1
y = 5x + 4
y=x+6
y = 4x – 1
y = -x + 9
y = 7x – 5
2y = 4x + 7
Divide by 2
Y = 2x + 3.5
3y = 7x – 6
Divide by 3
7
y = x2
3
2y = 8x + 5
2
3
1
0.5
y-intercept (c
or b)
5
7
-3
1
2
3.5
7
3
-2
-1
5
2y = 6x – 1
3y = 4x + 3
5y = 2x + 3
y+x=5
y =-x + 5
y + 3x = 7
2y – 5x = 3
Page 4
LINEAR FUNCTIONS
FINDING GRADIENT
To find the gradient (m) of a line joining two points A( x1 , y1 ) and B( x2 , y2 ) , use the
formula
m=
𝒚𝟏 −𝒚𝟐
𝒙𝟏 −𝒙𝟐
Find the gradient (m) of the line joining the points.
Points
(2,3) and (4, 7)
(3, 6) and (1, -4)
Gradient (m)
3−7
−4
m = 2−4 = −2= 2
m=
(1,2) and (5, 6)
(1, 3) and (3, 9)
(-1,6) and (3, 10)
(-2, 5) and (3, 10)
(0,2) and (3, 5)
(2, -3) and (1, 0)
(1,2) and (8, 10)
(5, 4) and (3, -2)
Page 5
6  (4) 10
=
=5
3 1
2
LINEAR FUNCTIONS
EQUATION OF LINE : GIVEN GRADIENT AND Y-INTERCEPT
The equation of a line is given by y = mx + c, where m is the gradient and c is the y–
intercept. To find the equation of a line we need
1. the gradient
2. the y-intercept
Example 1
Find the equation of the line having gradient 5 and crossing the y-axis at 3.
Solution: y = 5x + 3
Example 2
Find the equation of the line having slope -6 and y-intercept 4.
Solution: y = -6x + 4
EXERCISE
1.
Find the equation of the line having
gradient 7 and crossing the y-axis at 2.
2. Find the equation of the line having
gradient 3 and crossing the y-axis at -2.
3.
Find the equation of the line having
slope -1 and y-intercept 6.
4. Find the equation of the line having
gradient 9 and crossing the y-axis at 7.
5.
Find the equation of the line having
slope -2 and y-intercept 10.
6. Find the equation of the line having
slope -5 and y-intercept 1.
7.
Find the equation of the line having
gradient 1 and crossing the y-axis at
(0,5).
8. Find the equation of the line having
gradient 4 and crossing the y-axis at
(0,-3).
Page 6
LINEAR FUNCTIONS
Equation of a line : GRADIENT GIVEN AND A POINT
Example
Find the equation of the line having gradient 2 and passing through A(3,5).
Solution: Let 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) be the equation of the line
m = 2 x1 = 3 and y1 = 5
𝑦 − 5 = 2(𝑥 − 3)
𝑦 − 5 = 2𝑥 − 6
𝑦 = 2𝑥 − 6 + 5
𝑦 = 2𝑥 − 1
EXERCISE
1.
Find the equation of the line having
gradient 3 and passing through (4,5).
2. Find the equation of the line having
gradient 5 and passing through (-2,4).
3.
Find the equation of the line having
slope 6 and passing through (-1,3).
4. Find the equation of the line having
gradient -4 and passing through (2,1).
5.
Find the equation of the line having gradient -5 and passing through (-2,3).
Page 7
LINEAR FUNCTIONS
HOW TO SHOW A POINT LIES ON A LINE?
To show a point lies on a given line,
 replace the x and y value in the equation of the line
 we have to obtain the constant value
Examples
Show that the point A(2,5) lies on the
line with equation 4x + y = 13.
Solution
Replace x = 2 and y = 5 in the equation.
We have to obtain the right hand side
number.
4(2) + (5) = 13 (shown)
Does the point B(3,-2) lie on the line with
equation 2x + y = 10?
Solution
Replace x = 3 and y = -2 in the equation.
2(3) + (-2) = 4  10
Therefore B does not lie on the line.
The point A(p,3) lies on the line 4x – 2y = 2, find the value of p.
Solution
Replace x by p and y by 3 in the given equation and solve for p
4(p) – 2(3) = 2
4p – 6 = 2
4p = 8
p=2
EXERCISE
1. Show that the point (4,3) lies on the line 2. Show that the point (5,-1) lies on the
with equation 3x + 2y = 18.
line with equation x + 2y = 3.
3. Does the point (2,5) lie on the line with
equation 3x - y = 1?
4. The point A(t,4) lies on the line
4x – 2y = 12, find the value of t.
5. The point A(4,m) lies on the line
5x + 2y = 6, find the value of m.
6. The point A(q,4) lies on the line
3x – 5y = -2, find the value of q.
Page 8
LINEAR FUNCTIONS
DISTANCE BETWEEN TWO POINTS
To find the distance (d) between two points A( x1 , y1 ) and B( x2 , y2 ) , use the formula
d = √(𝑥1 − 𝑥2 )2 + (𝑦1 − 𝑦2 )2
Example
Find the distance between the points A(5,7) and B(2,3).
Solution
D=
( x 2  x1 ) 2  ( y 2  y1 ) 2
= √(5 − 2)2 + (7 − 3)2
= √32 + 42
= 5 units
EXERCISE
1. Find the distance between the points
A(1,2) and B(5,5).
2. Find the distance between the points
P(-1,2) and Q(5,10).
3. Find the distance between the points
A(-2,3) and B(10,8).
4. Find the distance between the points
L(0,-3) and M(24,4).
5. Find the distance between the points
A(1,4) and B(3,6).
6. Find the distance between the points
P(-2,-3) and Q(4,5).
Page 9
LINEAR FUNCTIONS
PARALLEL LINES
When two lines are parallel, they have the same gradient. For example,
y = 2x, y = 2x - 3 and y = 2x + 5 are parallel. We can also say that these lines belong to
the family of lines having gradient 2.
Similarly, y = 3x and 2y = 6x -1 are parallel because both lines have a slope of 3.
CLASS ACTIVITY 1
Two lines from each of these groups are parallel. Name them.
Group 1
A : y  3x  2
Gradient
Group 2
P : y  5x  2
Gradient
Group 3
L : y  4x  2
C : 3y  9x  5
Q : 2 y  12 x  1
R : 3 y  12 x  5
M : 2 y  8x  1
N : 3 y  15 x  1
D : y  2x  4
S : y  5x  7
P : y  4x  3
B : 2 y  8x  1
Gradient
Group 1 :
Group 2 :
Group 3 :
CLASS ACTIVITY 2
Which of the following are parallel? There may be more than two lines parallel.
Group 4
A : y  2x  3
B : 2 y  4x  1
C : 3y  6x  5
D : y  2x  9
Gradient Group 5
P: y  x2
Q : 2 y  4x  1
R : 3 y  3x  4
S:yx3
Gradient
Group 4 :
Group 5 :
Group 6 :
Page 10
Group 6
L : y  7x  2
M : 2 y  14 x  1
N : 3 y  21x  1
P : y  7x  0
Gradient
LINEAR FUNCTIONS
Equation of a line passing through a point and parallel to a given line
Example 1
Find the equation of the line passing through A(3,5) and parallel to y = 2x + 7.
Solution:
The gradient of the given line is 2.
Therefore, the gradient of the required line will be 2 as well.
Let 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) be the equation of the line
𝑦 − 5 = 2(𝑥 − 3)
𝑦 − 5 = 2𝑥 − 6
𝑦 = 2𝑥 − 6 + 5
∴ 𝑦 = 2𝑥 − 1
EXERCISE
1. Find the equation of the line passing
through (2,3) and parallel to 𝑦 = 4𝑥 + 11.
2. Find the equation of the line passing
through (-4,5) and parallel to 𝑦 = 3𝑥 − 7.
3. Find the equation of the line passing
through (1,-2) and parallel to 𝑦 = 5𝑥 + 8.
4. Find the equation of the line passing
through (0,6) and parallel to 2𝑦 = 6𝑥 − 7.
5. Find the equation of the line passing
through (1,5) and parallel to 3𝑦 = 6𝑥 − 5.
6. Find the equation of the line passing
through (-2,7) and parallel to 𝑦 − 5𝑥 = 4.
Page 11
LINEAR FUNCTIONS
Perpendicular lines
When two lines are perpendicular, the product of their gradients is -1.
1
1
For example, y = 2x and y =  x + 5 are perpendicular because 2 × − 2 = −1
2
3
4
3
4
Similarly, y = x + 1 and y =  x + 5 are perpendicular because 4 × − 3 = −1
3
4
CLASS ACTIVITY
Copy and complete the following table.
Equation of line
Gradient of
parallel line
y = 4x
m=4
3
x6
5
y = 3x
m=
y=
3
5
y = 5x – 3
y= 
2
x6
5
y = -6x + 5
y =  7x  2
y=
3
x 1
8
y =  x3
5
y =  x2
3
Page 12
Gradient of
perpendicular line
(flip and change sign)
1
m=4
5
m=3
LINEAR FUNCTIONS
How to show two lines are perpendicular?
To show two lines are perpendicular, simply multiply the gradients of both lines and the
result must be -1.
Examples
Show that the line L1 : 𝑦 = 4𝑥 + 5 and the
−1
line L2 : 𝑦 = 4 𝑥 + 3 are perpendicular.
Show that the lines 2𝑦 = 6𝑥 + 1 and 3𝑦 +
𝑥 = 7 are perpendicular.
Solution
Solution
Gradient of L1 = 4
Gradient of 2𝑦 = 6𝑥 + 1 is 3
1
Gradient of 3𝑦 + 𝑥 = 7 is − 3
1
Gradient of L2 = − 4
Product of gradients = 4 ×
−1
4
= −1 (𝑠ℎ𝑜𝑤𝑛)
Product of gradients = 3 ×
−1
3
= −1 (𝑠ℎ𝑜𝑤𝑛)
Are the lines 𝑦 = 5𝑥 − 3 and 2𝑦 = 𝑥 + 4 are perpendicular?
Solution
Gradient of 𝑦 = 5𝑥 − 3 is 5
Gradient of 2𝑦 = 𝑥 + 4 is
1
2
1
Product of gradients = 5 × 2 ≠ −1
Hence the two lines are NOT perpendicular.
EXERCISE
1. Show that the line L1 : 𝑦 = 2𝑥 + 5 and
−1
the line L2 : 𝑦 = 2 𝑥 + 3 are
perpendicular.
2. Show that the lines 2𝑦 = 8𝑥 − 1 and
4𝑦 + 𝑥 = 5 are perpendicular.
3. Are the lines 𝑦 = 2𝑥 − 3 and 2𝑦 = 𝑥 + 5
are perpendicular?
4. Are the lines 2𝑦 = 5𝑥 + 1 and 5𝑦 = −2𝑥
are perpendicular?
Page 13
LINEAR FUNCTIONS
Equation of a line passing through a point and perpendicular to a given line.
Example 1
Find the equation of the line passing through A(6,5) and perpendicular to y = 2x + 7.
Solution:
2
The gradient of the given line is 2 or 1.
Therefore the gradient of the required line will be –
1
. (flip and change sign)
2
Let 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) be the equation of the line
1
𝑚 = − 2 , 𝑥1 = 6 𝑎𝑛𝑑 𝑦1 = 5
1
𝑦 − 5 = − (𝑥 − 6) expand the bracket
2
1
𝑦 − 5 = −2𝑥 + 3
add 5 on both sides
1
∴ 𝑦 = −2𝑥 + 8
EXERCISE
1. Find the equation of the line passing
through A(4,3) and perpendicular to
𝑦 = 4𝑥 + 11.
2.
Find the equation of the line passing
through A(9,-5) and perpendicular to
3
𝑦 = − 4 𝑥 + 8.
3. Find the equation of the line passing
through A(10,7) and perpendicular to
5
𝑦 = 3 𝑥 + 4.
4.
Find the equation of the line passing
through A(6,-5) and perpendicular to
2𝑦 = 6𝑥 + 1.
Page 14
LINEAR FUNCTIONS
Equation of a line passing through two points
To find the equation of a line passing through two points we find
y  y1
1. the gradient by using 2
x 2  x1
2. the y-intercept, by choosing any one of the two given points.
Example 1
Find the equation of the line passing through A(3,1) and B(5,9)
Solution:
y  y1
Since the gradient is not given, we use the formula m = 2
to find m.
x 2  x1
9−1
8
𝑚 = 5−3 = 2 = 4
From the two points A and B, we can choose any point as
the final answer would be same.
Let us choose A (3,1) to find the y-intercept c.
Let 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) be the equation of the line
m = 4 x1 = 3 and y1 = 1
𝑦 − 1 = 4(𝑥 − 3)
𝑦 − 1 = 4𝑥 − 12
𝑦 = 4𝑥 − 12 + 1
𝑦 = 4𝑥 − 11
If we choose B(5,9) instead, then
m = 4 x1 = 5 and y1 = 9
𝑦 − 9 = 4(𝑥 − 5)
𝑦 − 9 = 4𝑥 − 20
𝑦 = 4𝑥 − 20 + 9
𝑦 = 4𝑥 − 11
same as before
EXERCISE
1. Find the equation of the line passing
through A(2,5) and B(6,9).
2. Find the equation of the line passing
through A(-2,3) and B(1,9).
3. Find the equation of the line passing
through P(0,2) and B(2,8).
4. Find the equation of the line passing
through A(-3,1) and B(1,5).
Page 15
LINEAR FUNCTIONS
How to sketch a line?
To sketch a line, we can either use a table of values or the intercept-method.
Example 1
Sketch the line y = 2x + 3.
Solution
We make a table as under. Note: two pairs of values is enough.
x
y = 2x + 3
0
3
1
5
Plot these two co-ordinates on the Cartesian axes to sketch the line.
EXERCISE
a
Sketch the line y = x + 3.
b
Sketch the line y = 3x – 1.
y
-4
y
10
10
8
8
6
6
4
4
2
2
-2
2
4
6
8
10
x
-4
-2
2
-2
-2
-4
-4
Page 16
4
6
8
10
x
LINEAR FUNCTIONS
c
Sketch the line y = 4 – x.
d
Sketch the line y = x – 2.
y
y
10
10
8
8
6
6
4
4
2
-4
2
-2
2
4
6
8
10
x
-4
-2
-2
2
4
6
8
x
10
-2
-4
-4
e
Sketch the line y = 5 – 2x.
f
Sketch the line y = 3x + 2.
y
-4
g
y
10
10
8
8
6
6
4
4
2
2
-2
2
4
6
8
10
x
-4
-2
2
-2
-2
-4
-4
Sketch the line y = 8 – 2x.
h
8
10
x
y
10
10
8
8
6
6
4
4
2
2
-2
6
Sketch the line y = 5x + 2.
y
-4
4
2
4
6
8
10
x
-2
-4
-2
2
-2
-4
-4
Page 17
4
6
8
10
x
LINEAR FUNCTIONS
Horizontal and Vertical lines
Horizontal lines are of the type y = a, where a is any number.
Vertical lines are of the type x = b, where b is any number.
Example
x = -4
y
x=3
10
y=5
5
-10
-5
5
x
10
-5
y = -8
-10
EXERCISE
1. On the same set of axes sketch the line y = 4, y = -1, and x = 2.
y
10
8
6
4
2
-4
-2
2
-2
-4
Page 18
4
6
8
10
x
LINEAR FUNCTIONS
2. On the same set of axes sketch the line x = 3, x = -1, and y = 3.
y
10
8
6
4
2
-4
-2
2
4
6
8
x
10
-2
-4
3. On the same set of axes sketch the line x = 5, x = -4, and y = 7.
y
10
8
6
4
2
-4
-2
2
4
6
8
10
x
-2
-4
4. On the same set of axes sketch the line x = 9, x = -2, and y = 6.
y
10
8
6
4
2
-4
-2
2
-2
-4
Page 19
4
6
8
10
x
LINEAR FUNCTIONS
How to find the equation of a line given the sketch?
Example
Find the equation of the following line.
(a)
(b)
Solution
This line has a y-intercept of 2.
The gradient can be obtained by
using any two points on the line.
(0,2) and (-2,0)
02
2

1
m=
20 2
So the equation is y = x + 2
Alternately, we use gradient is rise
over run to find the equation.
Page 20
Solution
This line has a y-intercept of 6.
The gradient can be obtained by
using any two points on the line.
(0,6) and (2,0)
06 6

 3
m=
20
2
So the equation is y = -3x + 6
LINEAR FUNCTIONS
10
2
4
6
8
– 4
2
2– 4
4
6
8
10
2
EXERCISE
10
2
4
6
8
– 4
2
2– 4
4
6
8
10
2
Find the equations of the following lines
y
y
10
10
8
8
6
6
4
4
2
– 4
2
– 2
2
4
6
8
10 x
– 4
– 2
10
2
4
6
8
– 4
2
2– 4
4
6
8
10
2
10
2
4
6
8
– 4
2
2– 4
4
6
8
10
2
– 2
2
4
6
10 x
8
– 2
– 4
– 4
y
y
– 4
10
10
8
8
6
6
4
4
2
2
– 2
10
2
4
6
8
– 4
2
2– 4
4
6
8
10
2
2
4
6
8
10
– 2
x
10
2
4
6
8
– 10
2
4
6
8
2– 4
4
2
– 4
– 2
2
4
6
10 x
8
– 2
– 4
– 4
y
y
10
10
8
8
6
6
4
4
2
2
– 4
– 2
2
– 2
– 4
– 2
2
4
6
8
10
x
– 4
– 2
– 6
– 8
– 4
– 10
Page 21
4
x
LINEAR FUNCTIONS
FUNCTIONS
In this section, we are going to have a look at functions such as f(x), read as f of x. In
the first section, the reader would be able to understand and try to find the image of a
number.
Example
Given that f(x) = 3x + 5, find
(a) f(4)
(b) f(-6)
Solution
(a)
f(4) = 3(4) + 5 [replace the x by 4, and evaluate ]
= 12 + 5
= 17
(b) F(-6) = 3(-6) + 5
= -18 + 5
= -13
Example 2
Given that g ( x)  x 2  3x  5 , evaluate g(-2).
Solution
g (2)  (2) 2  3(2)  5 [Remember to always use brackets when you substitute]
=4+6+5
= 15
Example 3
Given that f ( x)  4 x  3 and g ( x)  15  2 x , find
(a) The value of a for which f(a) = 17
(b) Solve g(p) = 35
(c) Solve f(x) = g(x).
Solution
(a) Replace x by a in f(x) and equate to 17
4a – 3 = 17
4a = 20
a=5
(b) 15 – 2p = 35
-2p = 20
p = -10
(c) 4x  3  15  2x
4x  2x  15  3
6 x  18
x3
Page 22
LINEAR FUNCTIONS
EXERCISE
1. Given that f ( x)  2 x  7 , find f(3).
2.
Given that f ( x)  3x  10 , find f(-2).
3. Given that f ( x)  13  4 x , find f(4).
4.
Given that f ( x)  x 2  4 , find f(-5).
5. Given that g ( x)  2 x 2  3x  1 , find g(2).
6.
Given that g ( x)  3x 2  4 x  5 , find g(-2).
7. Given that g ( x)  (2 x  1) 2 , find g(-4).
8.
Given that g ( x)  3( x  4) 2  5 , find g(5) .
9. Given f ( x)  10  2( x  3) 2 , find f(-2).
10. Given that f ( x)  2 x 2  5 , find f(3).
Page 23
LINEAR FUNCTIONS
11. Given f ( x)  2 x  1 and g ( x)  10  3x ,
find
12. Given f ( x)  5 x  4 and g ( x)  18  2 x ,
find
(a) The value of a for which f(a) = 13.
(a) The value of a for which f(a) = 24.
(b) Solve g(p) = 22
(b) Solve g(p) = 28
(c) Solve f(x) = g(x).
(c) Solve f(x) = g(x).
13. Given f ( x)  x 2  5 and g ( x)  5 x  1,
find
14. Given that f ( x)  x 2  3 and
g ( x)  7 x  2 , find
(a) The values of a for which
f(a) = 20.
(a) The values of a for which
f(a) = 39.
(b) Solve g(t) = 11
(b) Solve g(q) = 19.
Page 24
LINEAR FUNCTIONS
USING TECHNOLOGY TO SOLVE SIMULTANEOUS EQUATIONS
To solve a pair of simultaneous equations, use the following steps:



Main
Keyboard
2D








Insert the first equation in the first box, the second equation in the second box and
x,y in the box outside.
Exercise
Solve the following equations using your calculator.
(a) 𝑦 = 2𝑥 + 5 𝑎𝑛𝑑 𝑦 = 3𝑥 − 1
Answer : x = ……… and y = …………
(b) 𝑦 = 𝑥 + 4 𝑎𝑛𝑑 𝑦 = 4𝑥 − 8
Answer : x = ……… and y = …………
(c) 𝑥 + 𝑦 = 10 𝑎𝑛𝑑 2𝑥 − 𝑦 = 8
Answer : x = ……… and y = …………
(d) 3𝑥 − 𝑦 = 10 𝑎𝑛𝑑 2𝑥 + 5𝑦 = 1
Answer : x = ……… and y = …………
(e) 4𝑥 + 3𝑦 = 11 𝑎𝑛𝑑 5𝑥 − 𝑦 = 9
Answer : x = ……… and y = …………
(f) 𝑥 + 𝑦 = 9 𝑎𝑛𝑑 2𝑥 − 𝑦 = 6
Answer : x = ……… and y = …………
Page 25