Here is a quick recap of what we know about straight lines on the coordinate plane:
Any straight line on the coordinate plane is defined entirely by its gradient and its $y$yintercept. These two features are all we need to write an equation for the line.
The variable $m$m is used to represent the gradient, and the variable $c$c is used for the $y$yintercept.
We can represent the equation of any straight line, except vertical lines, using what is known as the gradientintercept form of a straight line.
$y=mx+c$y=mx+c
To represent a particular line on the coordinate plane, with it's own unique gradient and $y$yintercept, we simply replace $m$m and $c$c with their corresponding values.
We can use the applet below to see the effect of varying $m$m and $c$c on both the line and its equation.

As can be seen from the applet, we can make the following statements about the gradient.
The value of the gradient, $m$m, relates to the line as follows:
In algebra, any number written immediately in front of a variable, is called a coefficient. For example, in the term $3x$3x, the coefficient of $x$x is $3$3. Any number by itself is known as a constant term.
In the gradientintercept form of a line, $y=mx+c$y=mx+c, the gradient, $m$m, is the coefficient of $x$x, and the $y$yintercept, $c$c, is a constant term.
If we know the gradient, $m$m, and $y$yintercept, $c$c, of a line, we can substitute these values into $y=mx+c$y=mx+c to write the equation of the line.
Write the equation of a line that has a gradient of $\frac{3}{4}$34 and a $y$yintercept of $2$−2.
Solution
Substitute $m=\frac{3}{4}$m=34 and $c=2$c=−2 into $y=mx+c$y=mx+c
$y$y  $=$=  $mx+c$mx+c 
$y$y  $=$=  $\frac{3}{4}x+\left(2\right)$34x+(−2) 
$y$y  $=$=  $\frac{3}{4}x2$34x−2 
Write down the equation of a line which has a gradient of $3$−3 and crosses the $y$yaxis at $1$1.
Give your answer in gradientintercept form.
State the gradient and $y$yintercept of the equation $y=9x3$y=−9x−3.
Gradient  $\editable{}$ 
$y$yintercept  $\editable{}$ 
To graph a line from an equation, we use the gradient, $m$m, and the $y$yintercept, $c$c.
We begin by locating the $y$yintercept as a point on the $y$yaxis.
From this point, we can use the gradient to draw the correct slope of the line, as outlined in the three examples below:
Graph the line that has a gradient of $4$4 and a $y$yintercept of $1$−1.
Solution
By comparing the equation of the line with $y=mx+c$y=mx+c, we see that the gradient is $4$4 and the yintercept is $1$−1. The gradient ($4=\frac{4}{1}$4=41) tells us that for a 'run' of $1$1, we have a 'rise' of $4$4.
We can now create the graph on a coordinate plane, in a series of steps:
Graph the line with equation $y=\frac{2}{3}x$y=−23x.
Solution
By comparing the equation of the line with $y=mx+c$y=mx+c, we see that the gradient is $\frac{2}{3}$−23 and the yintercept is $0$0, meaning the line passes through the origin. The gradient tells us that for a 'run' of $3$3, we have a 'rise' of $2$−2.
We can now create the graph on a coordinate plane, in a series of steps:
Sketch a graph of the linear equation $y=4x+3$y=4x+3.
A horizontal line has a gradient of zero ($m=0$m=0), so the equation of the line becomes:
$y=c$y=c
where $c$c is the $y$yintercept of the line.
Here are two examples of horizontal lines:
A vertical line has an undefined gradient, so we can't use the gradientintercept form to write its equation. Instead we define a vertical line as having the equation:
$x=b$x=b
where $b$b is the $x$xintercept of the line.
Here are two examples of vertical lines:
Write down the equation of the graphed line.
represents information in symbolic, graphical and tabular form