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

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Daniel I

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4 March 2023

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Finding Equations of Straight Lines

Why do we want to know about straight lines and their equations?

Straight Line Graphs (Linear Graphs) have lots of uses in mathematics – one use is in navigation
We may want to know the equation of a straight line so we can program it into a computer that will plot the line on a screen, along with several others, to make shapes and graphics

How do we find the equation of a straight line?

The general equation of a straight line is $y = m x + c$ where;
- $m$ is the gradient,
- $c$ is the $y$ -axis intercept (or simply, the $y$ -intercept)
To find the equation of a straight line you need TWO things:
1. the gradient, $m$ , which you can put straight into $y = m x + c$
  - get this from the question directly, or from a diagram
2. any point on the line- substitute this point into $y = m x + c$ (as you already know $m$ ) and solve to find $c$
  - if given two points which you used to find the gradient, just choose either one of them for the point to find $c$
You may be asked to give the equation in the form $a x + b y + c = 0$
(especially if $m$ is a fraction)
If in doubt, SKETCH IT!

What if the line is not in the form y=mx+c?

A line could be given in the form $a x + b y + c = 0$
- It is harder to identify the gradient and intercept in this form
We can rearrange the equation into $y = m x + c$ , so it is easier to identify the gradient and intercept
- $a x + b y + c = 0$
- This is easiest done when the values of $a, b$ and $c$ are known
  - e.g. The equation $2 x + 5 y - 6 = 0$ can be rearranged to
    $5 y = - 2 x + 6 y = - \frac{2}{5} x + \frac{6}{5}$
  - So the gradient is $- \frac{2}{5}$ and the line intercepts the y-axis at the point $(0, \frac{6}{5})$

Worked example

(a)

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

We know that the gradient is 3 so the line takes the form

$y = 3 x + c$

To find the value of c, substitute (5, 4) into the equation

$4 = 3 (5) + c 4 = 15 + c c = - 11$

Replace c with −11 to complete the equation of the line

y = 3x − 11

(b)

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

First find m, the gradient
Identify two points the line passes through and work out the rise and run

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

cie-igce-core-rn-finding-equations-of-straight-lines-we-solution-1

∴ rise = 4 (4 - 0)
run = 8 (10 - 2)

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

!! Remember we also need to consider whether the line is "uphill" or "downhill" to find the gradient !!
The slope is downward in the left to right sense, so it is a negative gradient

∴ gradient, $m = - \frac{1}{2}$

We now know that the line takes the form

\begin{array}{rcl} y & = & - \frac{1}{2} x + c \end{array}

To find the value of c, substitute one of the points identified earlier into this equation
Here we've used (10, 0) as 0's often make the maths easier

$0 = - \frac{1}{2} (10) + c 0 = - 5 + c c = 5$

Replace c with 5 to complete the equation of the line

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

We can check against our sketch that this equation looks correct- it has a negative gradient and it crosses the y-axis at 5.

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Drawing Linear Graphs

How do we draw the graph of a straight line from an equation?

Before you start trying to draw a straight line, make sure you understand how to find the equation of a straight line – that will help you understand this
How we draw a straight line depends on what form the equation is given in
There are two main forms you might see:
y = mx + c and ax + by + c = 0
Different ways of drawing the graph of a straight line:

From the form y = mx + c
(you might be able to rearrange to this form easily)

plot c on the y-axis
go 1 across, m up (and repeat until you can draw the line)
From ax + by + c = 0

put x = 0 to find y-axis intercept

put y = 0 to find x-axis intercept
(You may prefer to rearrange to y = mx + c and use above method)
Work out and plot two points on the line

Pick a value of x, substitute this into the equation of the line (whichever form it is in) to find the corresponding y coordinate

Repeat this for a different value of x

Plot the two points and join them up

It is often a good idea to pick and plot a third point, as a check
All three points will be on the line if they're correct - if not, at least one of the points is incorrect!

How do we draw the graph of a straight line from a table of values?

Some questions will not give you a diagram, nor tell you a graph is a straight line (linear) graph
Instead they will give you a table of x and y values to complete using the equation of the straight line
- The line is then drawn using the values in the table as coordinates
If you recognise that the equation (or pattern in the values) will give a straight line graph then you will know if any points are incorrect when you plot them
For example, complete the table of value below for the equation $y = - 2 x + 6$

$x$ -3 -2 -1 0 1 2 3

$y$ 2 4

Substitute each x value in turn to find the corresponding y value

$y = - 2 \times (- 3) + 6 = 6 + 6 = 12$

and so on for the other missing y values
Then you would plot each coordinate and join them up to draw the straight line graph.
If any coordinates do not lie on the line you should go back and check your calculations for that one.

Examiner Tip

If plotting a straight line from a table of values then there is no need to plot them all
- You'll need to plot at least 2 points
- We recommend you plot a third too as a way of checking the first two
To complete a table of values, you can use the TABLE feature/mode of your calculator, if it has one
- Most modern calculators have this feature but it may be hidden under on screen menus rather than having a dedicated key

Worked example

On the axes below, draw the graphs of $y = 3 x - 1$ and $3 x + 5 y = 15$ .

For $y = 3 x - 1$ , first plot c, which is (0, −1)

Then, as m = 3, $\frac{rise}{run} = \frac{3}{1}$ . So plot a point 3 up and 1 right. Repeat at least once more and then join the points with a straight line. Extend the line to the edges of the grid.

The steps for $3 x + 5 y = 15$ are the same, but first we need to rearrange into the form y = mx + c

$\begin{array}{rcl} 5 y & = & - 3 x + 15 \\ y & = & - \frac{3}{5} x + 3 \end{array}$

Now we can plot c, which is (0, 3)

For the gradient, $\frac{rise}{run} = \frac{- 3}{5}$ . So plot a point 3 down and 5 right. There isn't space on the grid to repeat this so join the points with a straight line. Extend the line to the edges of the grid.

y=3x-1 and 3x+5y=15, IGCSE & GCSE Maths revision notes

Parallel Lines

What are parallel lines?

Parallel lines are lines that have the same gradient, but are not the same line
Parallel lines do not intersect with each other
You can easily spot that two lines are parallel when they are written in the form $y = m x + c$ , as they will have the same value of $m$ (gradient)

$y = 3 x + 7$ and $y = 3 x - 4$ are parallel
$y = 2 x + 3$ and $y = 3 x + 3$ are not parallel
$y = 4 x + 9$ and $y = 4 x + 9$ are not parallel; they are the exact same line

How do I find the equation of a line parallel to another line?

As parallel lines have the same gradient, a line of the form $y = m x + c$ will be parallel to a line in the form $y = m x + d$ , where $m$ is the same for both lines

If $c = d$ then they would be the same line and therefore not parallel

If you are asked to find the equation of a line parallel to $y = m x + c$ , you will also be given some information about a point that the parallel line, $y = m x + d$ passes through; $(x_{1}, y_{1})$
You can then substitute this point into $y = m x + d$ and solve to find $d$

Worked example

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

As the gradient is the same, the line that is parallel will be in the form:

$y = 3 x + d$

Substitute in the coordinate that the line passes through:

$1 = 3 (2) + d$

Simplify:

$1 = 6 + d$

Subtract 6 from both sides:

$- 5 = d$

Final answer:

$y = 3 x - 5$

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Previous:Coordinate GeometryNext:Types of Graphs

$x$	-3	-2	-1	0	1	2	3
$y$		2				4

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

Revision Note

Why do we want to know about straight lines and their equations?

How do we find the equation of a straight line?

What if the line is not in the form y=mx+c?

How do we draw the graph of a straight line from an equation?

How do we draw the graph of a straight line from a table of values?

What are parallel lines?

How do I find the equation of a line parallel to another 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: Daniel I