This document contains 5 math problems involving factorizing expressions, solving equations, evaluating expressions for given values, expanding expressions, and finding the highest common factor. It also provides context on working with straight line graphs, including finding the gradient and y-intercept of a line from its equation, finding the gradient between two points, finding the midpoint and a point that divides a line segment in a given ratio, and finding the x- and y-intercepts of a line.

1. 3. Solve 8x + 2 = 2(x + 7)
2. Factorise 9mn2 − 12m2n
5.
5mn2 + 5my
10
when m = 3, n = -2, y = 5
4. Expand and simplify (2y + 3)(y + 2)
1. What is the HCF of 84 and 120?
Starter
Straight Line Graphs

2. Starter Straight Line Graphs
3. Solve 8x + 2 = 2(x + 7)
2. Factorise 9mn2 − 12m2n
5.
5mn2 + 5my
10
when m = 3, n = -2, y = 5
4. Expand and simplify (2y + 3)(y + 2)
3mn(3n – 4m)
8x + 2 = 2x + 14 6x = 12 x = 2
2y2 + 4y + 3y + 6 2y2 + 7y + 6
5(3)(−2)
2
+ 5(3)(5)
10
60 + 75
10
135
10
13.5
1. What is the HCF of 84 and 120?
84
2
42
2 21
7 3
120
12
5
2 2
10
2 3 4
84 = 2 × 2 × 3 × 7
120 = 2 × 2 × 2 × 3 × 5
HCF = 2 × 2 × 3
HCF = 12

3. The x-axis and the y-axis
intersect at the origin.
The coordinates of the
origin are (0, 0).
The lines of the grid are
numbered using positive
and negative integers as
follows.
O 1 2 3 4
–4 –3 –2 –1
1
2
3
4
–4
–3
–2
–1
x-axis
y-axis
origin

6. Understand that an equation of the
form y = mx + c corresponds to a
straight line graph

7. Form of a straight line
𝑦 = 𝑚𝑥 + 𝑐
Gradient
‘steepness or slope’
y-intercept
Where the graph crosses the y-axis
When x = 0, what is y?
Ways to find the gradient of a
line:
rise
run
=
y direction
x direction
y-intercept

8. Be able to find the gradient and
y-intercept from a given line, or
graph

9. Gradient & y-intercept
Example 1
What is the gradient and y-intercept of the line y = 7x – 3?
Gradient = 7 y-intercept = -3
Example 2
What is the gradient and y-intercept of the line 4y –
16x + 8 = 0
Rearrange the
equation so it is in
the form y = mx + c
4y – 16x + 8 = 0
+ 16x
4y + 8 = 16x
- 8
4y = 16x - 8
÷ 4
y = 4x - 2
Gradient = 4
y-intercept = -2

10. Try Find the gradient and y-intercept of the following lines:
a) y = 3x + 7 b) y = -4x - 3 c) y = 9 – 3x d) y – 2x = 5
e) 5y = 15x + 20 f) e) 6y + 12x = 24 g) 11y + 5x = -4 h) 6y – 3x + 36 = 0
gradient = 3
y-intercept =7
gradient = - 4
y-intercept = - 3
gradient = - 3
y-intercept = 9 gradient = 2
y-intercept = 5
y = 2x + 5
gradient = 3
y-intercept = 4
y = 3x + 4
gradient = - 2
y-intercept = 4
6y = - 12x + 24
y = - 2x + 4
gradient = −
5
11
y-intercept = −
4
11
11y = - 5x - 4
y = -
5
11
x -
4
11
gradient =
1
2
y-intercept = - 6
6y = 3x - 36
y =
1
2
x - 6

11. Look at the following examples of straight lines and work
out their gradients:
Ways to find the gradient of a
line:
rise
run
=
y−direction
x−direction
Finding the gradient of a line
L1
L2
L3
L1
3
2
y−direction
x−direction
= −
3
2
y−direction
x−direction
= −
3
6
3
6
= −
1
2
L3
y−direction
x−direction
=
5
2
5
2
If your line starts low
and ends high, the
gradient will be
positive.
x
y
If your line starts high
and ends low, the
gradient will be
negative.
x
y
L2

18. Be able to find the
gradient between two
points

19. Finding the gradient between 2 points
If you are given two coordinates: (x1 , y1) and (x2 , y2)
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
𝑦2−𝑦1
𝑥2−𝑥1
You can find the gradient of the line between these two points by finding the
difference in the y’s, and dividing this value by the difference of the x’s
Example 1
AB is a line segment where A(4 , 8) and B(10, 12).
Find the gradient of the line segment AB.
(x1 , y1) (x2 , y2)
(4 , 8) (10 , 12)
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
12 − 8
10 − 4 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
4
6
=
2
3

20. Try Find the gradient between the following coordinates:
a) (4 , 9) and (6 , 12) b) (9 , 9) and (14 , 3) c) (10 , 6) and (6 , 8) d) (1 , 7) and (8 , -8)
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
12 − 9
6 − 4
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
3
2
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
3 − 9
14 − 9
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
− 6
5
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
8 − 6
6 − 10
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
2
−4
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
−8 − 7
8− 1
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
− 15
7
Extension:
The line segment AB has a gradient of 3, where A(a , 6)
and B(12, 8).
Find the value of a.
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
𝑦2−𝑦1
𝑥2−𝑥1
3=
12 − a
8 − 6
3=
12 − a
2
6 =12 − a −6 = −a a = 6

22. y = mx + c
y = mx - 6
y =
6
3
x - 6
y = 2x - 6
Gradient of a line:
rise
run
=
y−direction
x−direction
y = mx + c
gradient
(slope)
y-intercept (where
the line crosses the y-
axis)
Example:
Find the equation of
the given line.

23. y = - x -2
y = - 2x
y = x+3
y =
1
2
𝑥

24. y = - 2x -4
y = - 3x +1
y = x +4
y =
1
2
x -4

25. What is the equation of the straight line shown in the diagram?
y =-4

26. What is the equation of the straight
line shown in the diagram?
x =-3
What is the equation of the straight line
shown in the diagram?
y =3x-5

27. Be able to find the
midpoint between
two points

28. Finding the midpoint between 2 points
If you are given two coordinates: (x1 , y1) and (x2 , y2)
you can find the midpoint of the line between these two points by:
- adding the x-coordinates, then halve them
- adding the y-coordinates, then halve them
x1 + x2
2
,
y1 +
y2
2
(x1 , y1) (x2 , y2)
A(6 , 12) and B(4, 10).
x1 + x2
2
,
y1 +
y2
2
→
6 + 4
2
,
12 + 10
2
→
10
2
,
22
2
→ 5, 11
Example 1
AB is a line segment where A(6 , 12) and B(4, 10).
Find the midpoint of the line segment AB.

29. Answer (2, 2)
What is the midpoint of the straight line
segment joining the points (-2, 5) and
(6, -1)?
What is the
midpoint of AB?
Answer (1, 4
1
2
)
What are the
coordinates of the
point that divides AB
in the ratio 1 : 3 ?
Answer (-3, 5)
Use P = (
𝑚𝑥2+𝑛𝑥1
𝑚+𝑛
,
𝑚𝑦2+𝑛𝑦1
𝑚+𝑛
)
With 𝑥1 = -6 , 𝑦1 =7, 𝑥2=6, 𝑦2=-1, m=1 ,n=3
Therefore p =
(
1 ×6+3×(−6)
1+3
,
1 × −1 +3 ×7
1+3
) = (
−12
4
,
20
4
)=(-3,5)

30. A is the point (-6, 5) and B is the point (10, -3)
P divides the straight line segment AB in the ratio 1 : 3 (i.e. it is a quarter of the way
between A and B)
What are the coordinates of P? Answer (-2, 3)
Use Q = (
𝑚𝑥2+𝑛𝑥1
𝑚+𝑛
,
𝑚𝑦2+𝑛𝑦1
𝑚+𝑛
)

31. What are the coordinates of the point that
divides CB in the ratio 3 : 1 ?
Answer (4, -2)
Use Q = (
𝑚𝑥2+𝑛𝑥1
𝑚+𝑛
,
𝑚𝑦2+𝑛𝑦1
𝑚+𝑛
)

32. Be able to find the
coordinate of a point
crossing either axis

33. Finding x and y interception points
Example 1
Line A has the equation y = 3x + 2.
a) What are the coordinates of the point where it crosses the y-axis?
b) What are the coordinates of the point where it crosses the x-axis?
To find an x-intercept, the y-coordinate
at this point is always 0.
To find a y-intercept, the x-coordinate
at this point is always 0.
a) y-intercept
x = 0 at this point
y = 3x + 2
y = 3(0) + 2
y = 0 + 2
y = 2
(0, 2)
b) x-intercept
y = 0 at this point
y = 3x + 2
0 = 3x + 2
-2 -2
-2 = 3x
÷ 2 ÷ 2
−
2
3
= x (−
2
3
, 0)

34. Find the y-intercepts of x2 + y2 + 2x − 5y − 24 = 0
To find the y- intercepts, set x = 0
x2 + y2 + 2x − 5y − 24 = 0
⇒ y2 − 5y − 24 = 0
Factor y2 − 5y − 24 = 0:
⇒ y2 − 8y + 3y − 24 = 0
⇒ y(y − 8) + 3(y − 8) = 0
⇒ (y + 3)(y − 8) = 0
Solve:
⇒ y = -3 or 8
Therefore the y-intercepts are (0, -3) and (0, 8)

35. Find the x-intercepts of 9x2 + 9y2 + 12x + 23y − 12 = 0
Make sure you find the x-intercepts and not the y-intercepts.
To find the x-intercepts, set y = 0
⇒ 9x2 + 12x − 12 = 0 ⇒ 3x2 + 4x − 4 = 0
⇒ 3x2 + 6x − 2x − 4 = 0 ⇒ 3x(x + 2) − 2(x + 2) = 0
⇒ (3x − 2)(x + 2) = 0
X=
2
3
or -2
Therefore, the x-intercepts are
(
2
3
, 0) (-2, 0)

36. What are the x-intercepts of
y = x3 + 64?
y = x3 + 64 can be factored using the "sum of
two cubes" formula:
m3 + n3 = (m + n)(m2 − mn + n2)
Put m = x and and n3 = 64, so n = 4:
y = x3 + 64 = (x + 4)(x2 − 4x + 16)
The second factor is quadratic with discriminant
equal to (-4)2 − 4 × 1 × 16 = 16 − 64 = -48
Since the discriminant is negative, it has no real
roots.
So the only intercept comes from the other factor:
(x + 4) = 0 ⇒ x = -4
So, The x-intercept is (-4, 0)
y = x2 − 3x + k has the same y-intercept
as y = x2 + 9x − 10
What are the x-intercepts of
y = x2 − 3x + k?
To find the y-intercept: set x = 0
Therefore, the y coordinate of the y-intercept of y
= x2 + 9x − 10 is -10
y = x2 − 3x + k has the same y-intercept, so k = -10
⇒ y = x2 − 3x − 10
= x2 + 2x − 5x − 10
= x(x + 2) − 5(x + 2)
= (x − 5)(x + 2)
To find the x-intercepts: set y = 0
⇒ (x − 5)(x + 2) = 0 ⇒ x = 5 or -2
So, the x-intercepts are (-2, 0) and (5, 0)

37. y = x2 − ax − b has x=intercepts at (-2, 0) and (3, 0)
What is its y-intercept?
(-2, 0) is an intercept
⇒ y = 0 when x = -2
⇒ 0 = (-2)2 − a × (-2) − b
⇒ 4 + 2a − b = 0 ... (1)
(3, 0) is an intercept
⇒ y = 0 when x = 3
⇒ 0 = 32 − a × 3 − b
⇒ 9 − 3a − b = 0 ... (2)
Subtract equation (1) from equation (2)
⇒ 5 - 5a = 0
⇒ a = 1
Substitute a = 1 into (1)
⇒ 4 + 2 − b = 0
⇒ b = 6
So, the equation of the curve is y = x2 − x − 6
And the y-intercept is (0, -6)

38. What are the intercepts of y = x2 + 4x − 12?
Answer (-6, 0) (2, 0) and (0, -12)
What are the intercepts of y = 8x2 + 2x − 15 ?
Answer (1.25, 0), (-1.5, 0) and (0, -15)
What are the x-intercepts of y = x3 + x2 − 4x − 4?
Answer (2, 0), (-2, 0) and (-1, 0)
What are the x-intercepts of y = 20x3 − 8x2 − 45x + 18?
Answer (1.5, 0), (-1.5, 0) and (0.4, 0)

39. Be able to form the
equation of a straight
line when given a
point (x,y)

40. Forming equations when given a point (x,y)
Remember: the standard format of a
straight line graph is y = mx + c
E.g. Find the equation of the straight
line that has a gradient of -2 and goes
through the point (-3, -10)
y = mx + c
Gradient (m) = -2
y = - 2x + c
(-3,-10) → x = -3, y = -10
-10 = -2(-3) + c
-10 = 6 + c
- 6 - 6
-16 = c
y = - 2x – 16
E.g. Find the equation of the straight
line that cuts the y-axis at 4 and goes
through the point (4, 7)
y = mx + c y-intercept (c) = 4
y = mx + 4
(4,7) → x = 4, y = 7
7 = m(4) + 4
7 = 4m + 4
- 4 - 4
3 = 4m
y =
3
4
x + 4
Sub into y = -2x + c to find c
Sub into y = mx + 4 to find m
m =
3
4

41. Be able to form the
“point –slope” form of
the equation of a
straight line.

42. The "point-slope" form of the equation of a
straight line is:
The equation is useful when we know:
•one point on the line: (x1,y1)
•and the slope of the line: m,
and want to find other points on the line.

46. Answer:
the slope, m = Rise/Run = -2/1 = -2
Now use y - y1 = m(x - x1)
Substitute x1 = 4, y1 = -3 and m = -2
⇒ y - (-3) = -2(x - 4)
⇒ y + 3 = -2(x - 4)

47. Answer:
This is a vertical line and every
point on the line has x
coordinate 3.
It's equation, therefore, is x = 3

48. Answer:

49. Answer:

50. Review Understand that an equation of the
form y = mx + c corresponds to a
straight line graph
Be able to find the gradient and y-
intercept from a given line, or
graph
Be able to find the equation of a
given straight line
Be able to form the equation of a
straight light when given a point
(x,y)
Be able to find the midpoint
between two points
Be able to find the gradient between
two points
Be able to find the coordinate of a
point crossing either axis
1. What is the gradient and y-intercept of the straight line 2y + 6x – 8 = 0
2. What is the gradient and y-intercept of the line y = 4 – 7x?
3. What is the gradient and y-intercept of the line y = 3(2x + 1)
4. Find the equation of the following straight line:
5. The line segment AB exists where A(3, 6) and B(10, - 3 )
a) Find the gradient of the line segment AB
b) Find the midpoint of the line segment AB
6. A line has gradient 6 and goes through the point (10, 12). What is the
equation of the straight line?
7. The line y = 7x + 3 cuts the x-axis at point P. Find the coordinate of P.

51. Answers
Understand that an equation
of the form y = mx + c
corresponds to a straight line
graph
Be able to find the gradient
and y-intercept from a given
line, or graph
Be able to find the equation of
a given straight line
Be able to form the equation of
a straight light when given a
point (x,y)
Be able to find the midpoint
between two points
Be able to find the gradient
between two points
Be able to find the coordinate
of a point crossing either axis
7. The line y = 7x + 3 cuts the x-axis at point P. Find the
coordinate of P.
1. What is the gradient and
y-intercept of the straight
line 2y + 6x – 8 = 0
2. What is the gradient
and y-intercept of the line
y = 4 – 7x?
3. What is the gradient and y-
intercept of the line y = 3(2x + 1)
4. Find the equation of the
following straight line:
5. The line segment AB exists where A(3, 6) and B(10, - 3 )
a) Find the gradient of the line segment AB
b) Find the midpoint of the line segment AB
6. A line has gradient 6 and goes through the point (10, 12).
What is the equation of the straight line?
2y = -6x + 8
y = -3x + 4
gradient = -3
y-intercept = 4
y = -7x + 4
gradient = -7
y-intercept = 4
y = 6x + 3
gradient = 6
y-intercept = 3
y = mx + c
gradient =
2
3
y-intercept = 1
y =
2
3
x + 1
a)
𝒚𝟐−𝒚𝟏
𝒙𝟐−𝒙𝟏
→
−𝟑−𝟔
𝟏𝟎−𝟑
→ −
𝟗
𝟕
b)
x1+x2
𝟐
,
y1+
y2
𝟐
3+10
2
,
6+−3
2
→(6.5, 1.5)
y = mx + c
y = 6x + c → 12 = 6(10) + c → 12 = 60 + c
→ - 48 = c → y = 6x - 48
0 = 7x + 3 → - 3 = 7x → −
3
7
=x → ( −
3
7
, 0)

52. Be able to find the
distance between
two points

53. Finding the distance between two points
To find the distance between two points,
use your understanding of Pythagoras’
Theorem
distance= 7 − 2 2+ 8 − 3 2
E.g.1 Find the distance between points
A and B where A( 2, 3) and B( 7, 8) to 2dp
A( 2, 3) and B( 7, 8)
(x1, y1) and (x2, y2)
= 5 2+ 5 2
= 25 + 25
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑥2 − 𝑥1
2 + 𝑦2 − 𝑦1
2
= 50
= 7.07(2dp)
1 2 3 4 5
6
7 8
0
0
1
2
3
4
5
7
8
6
5
5
Using Pythagoras’ theorem:
a2 + b
2
=c2
52 + 52=c2
25 + 25 =c2
50 =c2 50 = c c = 7.07(2dp)
Alternative Method

54. Try
Find the distance between the following coordinates:
a) (4 , 9) and (6 , 12) b) (9 , 9) and (14 , 3)
6 − 4 2+ 12 − 9 2
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑥2 − 𝑥1
2 + 𝑦2 − 𝑦1
2
c) (10 , 6) and (6 , 8) d) (1 , 7) and (8 , 3.5)
2 2+ 3 2
4 + 9
13
3.61cm (2dp)
14 − 9 2+ 3 − 9 2
5 2+ −6 2
25 + 36
61
7.81cm (2dp)
6 − 10 2+ 8 − 6 2
4 2+ 2 2
16 + 4
2 5
4.47cm (2dp)
8 − 1 2+ 3.5 − 7 2
7 2+ − 3.5 2
61.25
7.83cm (2dp)
49 + 12.25

55. Be able to complete a
table of values to plot
a straight line graph

56. E.g.
Draw the graph of y = 3x – 1 for values -2 < x < 2
Draw a table of values
x -2 -1 0 1 2
y
× 3
x
- 1
y
-7 5
2
- 1
- 4
This means your x-values begin
from -2 and end at 2 in your
table
1 2 3 4
-4 -3 -2 -1
8
7
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
-7
-8
Plot the values

57. Try Use the table of values below to plot the following graphs:
y = 2x – 3 y = - x + 3 y = 0.5x + 3
x -2 -1 0 1 2
y
x -2 -1 0 1 2
y -7 -5 -3 -1 1
x -2 -1 0 1 2
y 5 4 3 2 1
x -2 -1 0 1 2
y 2 2.5 3 3.5 4

58. Understand how the
gradients of parallel
lines are related
Understand how the
gradients of perpendicular
lines are related

59. Parallel Lines
What do you know about parallel lines?
What do you know about perpendicular lines? Perpendicular Lines
Same gradient
Opposite sign, reciprocal
To prove two lines are perpendicular, the product of their
gradients will be -1.
They never meet They have the same steepness
They meet at a right-angle. They meet at a right-angle.
the point where
two city roads
intersect
floor tiles
double yellow
lines
piano
keys
Train
tracks

60. Be able to prove that
two lines are parallel
or perpendicular

61. Prove that Line A and Line B are
parallel:
Line A: y = 3x + 4
Line B: 6y – 18x + 12 = 0
Line B
6y – 18x + 12 = 0
- 12 - 12
6y – 18x = - 12
+ 18x + 18x
6y = 18x - 12
÷ 4 ÷ 4
Line A and B are parallel as
they have the same gradient
(m = 3)
Prove that Line A and Line B are
perpendicular:
Line A: y = 2x + 4
Line B: 4y + 2x – 1 = 0
Line B 4y + 2x – 1 = 0
+ 1 + 1
4y + 2x = 1
- 2x - 2x
4y = - 2x + 1
y = -½ x + 1
2× −
1
2
= -1
Line A and Line B are perpendicular as
the product of their gradients is – 1.
÷ 6 ÷ 6
y = 3x - 2
To prove two lines are perpendicular, the
product of their gradients will be -1.

62. Generate equations of a line
parallel or perpendicular to
a straight line graph

63. Example:
Line A has the equation y = 2x + 1.
Line B is parallel to line A and goes through the point (3, 2).
What is the equation of line B?
y = mx + c
Parallel lines = same gradient
y = 2x + c
(3,2) is a point on this line
x = 3, y = 2
2 = 2(3) + c
2 = 6 + c
- 6 - 6
- 4 = c
So, y = 2x - 4
If A is in the form of a straight line, B must
also be a straight line as they are
both parallel!
To find the y-intercept (c) sub in the given
coordinate (x,y)
Don’t forget to sub c back in to give your answer in the form
y = mx + c!

64. Line A has the equation y = 5x + 2.
Line B is parallel to line A and goes through the
point (-1, -9). What is the equation of line B?
Line A has the equation y = x + 2
Line B is parallel to line A and goes through
the point (4, -1). What is the equation of line B?
Line A has the equation y = 0.5x + 2
Line B is parallel to line A and goes through the
point (4, - 1). What is the equation of line B?
Line A has the equation y = –2x – 1. Line B is
parallel to line A and goes through the point
(-4, - 3). What is the equation of line B?
y = mx + c
y = 5x + c
x = - 1, y = - 9
- 9 = 5(- 1) + c
- 9 = -5 + c
+ 5 + 5
- 4 = c
So, y = 5x – 4
y = mx + c
y = x + c
x = 4, y = - 1
- 1 = 4 + c
- 4 - 4
- 5 = c
So, y = x - 5
y = mx + c
y = 0.5x + c
x = 4, y = -1
-1 = 0.5(4) + c
-1 = 2+ c
- 2 - 2
- 3= c
So, y = 0.5x - 3
y = mx + c
y = - 2x + c
x = - 4, y = - 3
-3 = - 2(- 4) + c
-3 = 8 + c
- 8 - 8
- 11 = c
So, y = - 2x - 11

65. Review
1. The line segment AB has coordinates A(4,9) and B(13, 18).
What is the length of the line AB?
2. Complete the table of values and plot the graph of y = -2x – 4
3. Show that line A and line B are parallel:
Line A: 3y + 6x – 15 = 0 and Line B: y = -2x + 10
4. Prove that the following straight lines are perpendicular:
Line A: y = −
3
2
𝑥 + 5 and Line B: 3y – 2x + 18 = 0
5. Line A has the equation y = 3x – 5.
Line B is parallel to line A and goes through the point (3, 12).
What is the equation of the line?
6. Line A has the equation y = 10x + 4.
Line B is perpendicular to line A and goes through the point (2, 9).
What is the equation of the line?
Be able to find the distance between
two points
Be able to complete a table of values
to plot a straight line graph
Understand how the gradients of
parallel lines are related
Understand how the gradients of
perpendicular lines are related
Be able to prove that two lines are
parallel or perpendicular
Generate equations of a line
parallel or perpendicular to a
straight line graph
x -2 -1 0 1 2
y

66. Answers
1. The line segment AB has coordinates A(4,9) and B(13, 18).
What is the length of the line AB?
2. Complete the table of values and plot the graph of y = -2x – 4
3. Show that line A and line B are parallel:
Line A: Line A: 3y + 6x – 15 = 0 and Line B: y = -2x + 10
4. Prove that the following straight lines are perpendicular:
Line A: y = −
3
2
𝑥 + 5 and Line B: 3y – 2x + 18 = 0
5. Line A has the equation y = 3x – 5. Line B is parallel to line A and goes through the
point (3, 12). What is the equation of the line?
6. Line A has the equation y = 10x + 4. Line B is perpendicular to line A and goes
through the point (2, 9). What is the equation of the line?
Be able to find the distance between
two points
Be able to complete a table of values
to plot a straight line graph
Understand how the gradients of
parallel lines are related
Understand how the gradients of
perpendicular lines are related
Be able to prove that two lines are
parallel or perpendicular
Generate equations of a line
parallel or perpendicular to a
straight line graph
x -2 -1 0 1 2
y
13 − 4 2+ 18 − 9 2 = 12.73 (2dp)
0 -2 -4 -6 -8
Line A rearranged: y = -2x + 5
2
3
× −
3
2
= −
6
6
= -1
x
They are perpendicular as the
product of their gradients is -1.
Line B rearranged:
y =
2
3
x - 6
They are parallel as they have the same gradient
y = 3x + c → 12 = 3(3) + c → 12 = 9 + c → c = 3 → y = 3x + 3
y = −
1
10
x + c → 9 = −
1
10
(2) + c → 9 = - 0.2 + c → c = 9.2 → y = −
1
10
x + 9.2 or y = −
1
10
x + −
46
10