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Straight Line Graphs (y = mx + c) (CIE IGCSE Maths: Core)

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Mark

Last updated

15 February 2024

Finding Equations of Straight Lines

What is the equation of a straight line?

The general equation of a straight line is y = mx + c where
- m is the gradient
- c is the y-intercept
  - The value where it cuts the y-axis
y = 5x + 2 is a straight line with
- gradient 5
- y-intercept 2
y = 3 - 4x is a straight line with
- gradient -4
- y-intercept 3

How do I find the equation of a straight line from a graph?

Find the gradient by drawing a triangle and using
- - Positive for uphill lines, negative for downhill
Read off the y-intercept from the graph
- Where it cuts the y-axis
Substitute these values into y = mx + c

What if no y-intercept is shown?

If you can't read off the y-intercept
- find any point on the line
- substitute it into the equation
- solve to find c
For example, a line with gradient 6 passes through (2, 15)
- The y-intercept is unknown
  - Write y = 6x + c
- Substitute in x = 2 and y = 15
  - 15 = 6 × 2 + c
  - 15 = 12 + c
- Solve for c
  - c = 3
- The equation is y = 6x + 3

What are the equations of horizontal and vertical lines?

A horizontal line has the equation y = c
- c is the y-intercept
A vertical line has the equation x = k
- k is the x-intercept
For example
- y = 4
- x = -2

Worked example

(a)

Find the equation of the straight line shown in the diagram below.

Graph of a straight line with negative gradient

Find m, the gradient
Identify any two points the line passes through and work out the rise and run

Line passes through (2, 4) and (10, 0)

Finding the equation of a straight line from a graph

The rise is 4
The run is 8

Calculate the fraction $\frac{rise}{run}$

$\frac{rise}{run} = \frac{4}{8} = \frac{1}{2}$

The slope is downward (downhill), so it is a negative gradient

gradient, $m = - \frac{1}{2}$

Now find the y-intercept
The line cuts the y-axis at 5

y-intercept,

\begin{array}{rcl} c & = & 5 \end{array}

Substitute the gradient, m, and the y-intercept, c, into y = mx + c

$y = - \frac{1}{2} x + 5$

(b)

Find the equation of the straight line with a gradient of 3 that passes through the point (5, 4).

Substitute m = 3 into y = mx + c
Leave c as an unknown letter

$y = 3 x + c$

Substitute x = 5 and y = 4 into the equation
Solve the equation to find c

$\begin{array}{rcl} 4 & = & 3 \times 5 + c \\ 4 & = & 15 + c \\ - 11 & = & c \end{array}$

You now know c
Replace c with −11 to complete the equation of the line

y = 3x − 11

Drawing Linear Graphs

How do I draw a straight line from a table of values?

You may be given a table of values with no equation
Use the x and y values to form a point with coordinates (x , y )
- Then plot these points
- Use a ruler to draw a straight line through them
  - All points should lie on the line
For example
- The points below are (-3, 0), (-2, 2), ... etc

$x$	-3	-2	-1	0	1	2	3
$y$	0	2	4	6	8	10	12

How do I draw a straight line using y = mx + c?

Use the equation to create your own table of values
- Choose points that are spread out across the axes given
For example, plot y = 2x + 1 on axes from x = 0 to x = 10
- Substitute in x = 0, x = 5 and x = 10 to get y coordinates
  - Then plot those points

$x$	0	5	10
$y$	1	11	21

How do I draw a straight line without using a table of values?

Start at the y-intercept, c
Then, for every 1 unit to the right, go up m units
- where m is the gradient
- If m is negative, go down
This creates a sequence of points which you can then join up
- Be careful of counting squares if axes have different scales
  - 1 unit might not be 1 square

Examiner Tip

Always plot at least 3 points (just in case one of your end points is wrong!)

Worked example

On the same set of axes, draw the graphs of $y = 3 x - 1$ and $y = - \frac{3}{5} x + 3$ .

For $y = 3 x - 1$ , create a table of values

$x$	0	1	2
$y$	-1	2	5

Plot the points (0, -1), (1, 2) and (2, 5)
Connect with a straight line
(Alternatively, start at the y-intercept and add 3 for every 1 unit to the right)

For $y = - \frac{3}{5} x + 3$ , create a table of values
Because of the fraction, x = 5 is a good point to include

$x$	0	3	5
$y$	3	1.2	0

Plot both lines on the same set of axes

Plotting two straight lines on the same axes

Parallel Lines

What are parallel lines?

Parallel lines are straight lines with the same gradient
- Two parallel lines will never meet
  - They just stay side-by-side forever
The equation of the line parallel to y = mx + c is y = mx + d
- $y = 2 x + 1$ and $y = 2 x + 5$ are parallel
- $y = 2 x + 1$ and $y = 3 x + 1$ are not parallel

How do I find the equation of a parallel line?

For example, to find the equation of the line parallel to y = 2x + 1 which passes through the point (3, 14)
- write the parallel line as y = 2x + d
  - using the same gradient of 2
- substitute x = 3 and y = 14 into this equation
  - 14 = 2 × 3 + d
  - 14 = 6 + d
- solve to find d
  - d = 8
- The equation is y = 2x + 8

Worked example

Find the equation of the line that passes through the point (2, 1), which is parallel to the straight line $y = 3 x + 7$ .

Parallel means the gradient will be the same
Use y = mx + d where m = 3

$y = 3 x + d$

Substitute in x = 2 and y = 1

$1 = 3 \times 2 + d$

Simplify the equation

$1 = 6 + d$

Solve the equation to find d (by subtracting 6 from both sides)

$- 5 = d$

Write out the final answer in the form y = mx + d

$y = 3 x - 5$

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Straight Line Graphs (y = mx + c) (CIE IGCSE Maths: Core)

Revision Note

What is the equation of a straight line?

How do I find the equation of a straight line from a graph?

What if no y-intercept is shown?

What are the equations of horizontal and vertical lines?

How do I draw a straight line from a table of values?

How do I draw a straight line using y = mx + c?

How do I draw a straight line without using a table of values?

What are parallel lines?

How do I find the equation of a parallel line?

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Mark