Straight-Line-Graphs-Final.pptxpptpptppt

3. Solve 8x + 2 = 2(x + 7)
2. Factorise 9m
5. 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

Starter Straight Line Graphs
3. Solve 8x + 2 = 2(x + 7)
2. Factorise 9m
5. 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
4
2
2
2
1
7 3
12
0
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

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

Understand that an equation of the
straight line graph

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: y-intercept

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

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

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 =
y-intercept =
11y = - 5x - 4
y = - x -
gradient =
y-intercept = - 6
6y = 3x - 36
y = x - 6

Look at the following examples of straight lines and work
out their gradients:
Ways to find the gradient of a
line:
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

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

Write down the equation of each line
y = mx + c
y = - x + c
y = - x + c
y = - x + 4
y = mx + c
y = x + c
y =x - 2
Line end low?
Negative gradient
Line end high?
Positive gradient
Line end high?
Positive gradient
y = mx + c
y = - x + c
y = -x + 5
Avoid decimals in
your fractions –
just double the
fraction!

Be able to find the
gradient between two
points

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

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)
12 − 9
6 − 4
3
2
3 − 9
14 − 9
− 6
5
8 − 6
6 − 10
2
−4
− 8 − 7
8− 1
− 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.
𝑦2 − 𝑦1
𝑥2 − 𝑥1
3 =
8 − 6
12 − a 3 =
2
1 2 - a
3 6 -3 a = 2 −3a = − 34 a = 34 /3

Be able to find the
midpoint between
two points

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
(x 1+ x 2
2
,
y 1+ y 2
2 )
(x1 , y1) (x2 , y2)
A(6 , 12) and B(4, 10).
(x 1+x 2
2
,
y 1+y 2
2 )→ (6+ 4
2
,
1 2+10
2 )→
(1 0
2
,
2 2
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.

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

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
÷ 3 ÷ 3
−
2
3
= x
(, 0)

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

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 = x + 4
Sub into y = -2x + c to find c
Sub into y = mx + 4 to find m
m =

Review Understand that an equation of the
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.

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 =
y-intercept = 1
y = x + 1
a) b)
(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)

Be able to find the
distance between
two points

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)
(, ) and (, )
= √(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
+ b2
=c2
52
+ 52
=c2
25 + 25
5 √50= c c = 7.07(2dp)
Alternative Method

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

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

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

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

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

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
Real-life examples
They meet at a right-angle. They meet at a right-angle.
Real-life examples
the point where
two city roads
intersect
floor
tiles
double
yellow lines
piano
keys
Train
tracks

Be able to prove that
two lines are parallel
or perpendicular

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
= -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.

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

Example:
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!

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

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 = + 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
perpendicular lines are related
Be able to prove that two lines are
parallel or perpendicular
parallel or perpendicular to a
straight line graph
x -2 -1 0 1 2
y

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 = + 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
parallel lines are related
perpendicular lines are related
Be able to prove that two lines are
parallel or perpendicular
parallel or perpendicular to a
straight line graph
x -2 -1 0 1 2
y
= 12.73 (2dp)
0 -2 -4 -6 -8
Line A rearranged: y = -2x + 5
= -1
x
They are perpendicular as the
product of their gradients is -1.
Line A rearranged:
y = 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 = x + c 9 = (2) + c 9 = - 0.2 + c c = 9.2 y = x + 9.2 or y = x +

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