The linear function is one of the most important in mathematics. Its graph is a straight line and is a simple way of representing data. An understanding of linear functions and their graphs is an essential requirement for many courses in science, engineering and mathematics. This module deals with linear graphs, in particular:

• Graphs of straight lines;
• The [latex]x[/latex] and [latex]y-[/latex] intercepts of a linear graph;
• The gradient of a straight line;
• The equation of a straight line;
• Equation of a line given a point on the line and its gradient;
• Equation of a line given two points.

Do you need this chapter?

Here is a short quiz to check your understanding of the material in this Chapter.

Graphs of linear functions

Graphs of linear functions are straight lines. Consider the graph in Fig. 7.1.

Some important features of the graph are:

• The point [latex]\left(b,0\right)[/latex] where the line cuts the [latex]x-[/latex] axis. This is called the [latex]x-[/latex] intercept.
• The point [latex]\left(0,c\right)[/latex] where the line cuts the [latex]y-[/latex] axis. This is called the [latex]y-[/latex] intercept. Often we refer to the number [latex]c[/latex] as the [latex]  y-[/latex] intercept.
• The slope of the graph. This is also called the gradient [latex]m[/latex] and is defined as

\begin{align}m&=\frac{\text{rise}}{\text{run}}\end{align}

where the rise and run are defined in Fig. 7.1. Note that for a straight line, the gradient is a constant.

• A line is defined by two points or a point on the line and the gradient of the line.

Examples: Straight line properties

1. What are the [latex]x[/latex] and [latex]y-[/latex] intercepts of the line in Fig. 7.2 ?

 

Solution:

The [latex]x-[/latex]intercept is [latex]\left(1,0\right)[/latex].

The [latex]y-[/latex]intercept is [latex]\left(0,-2\right)[/latex] or simply [latex]-2[/latex].

 

2. What are the [latex]x[/latex] and [latex]y-[/latex] intercepts of the line in Fig. 7.3 ?

 

Solution:

The [latex]x-[/latex] intercept is [latex]\left(-1,0\right)[/latex]. The [latex]y-[/latex] intercept is [latex]\left(0,-1\right)[/latex] or simply [latex]-1[/latex].

The Gradient of a straight line

The slope of the straight line is called the gradient denoted [latex]m[/latex]. This is the ratio of the vertical change (rise) to the horizontal change (run) between two points [latex]\left(x_{1},y_{1}\right)[/latex] and [latex]\left(x_{2},y_{2}\right)[/latex] on the line as shown in Fig. 7.4.

From Fig. 7.4 we see that [latex]x_1=0.4[/latex], [latex]x_2=2.4[/latex], [latex]y_1=1.2[/latex] and [latex]y_2=2.2[/latex]. The gradient of the line,

\begin{align*}
m& =\frac{\text{rise}}{\text{run}}\\
&=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\
&=\frac{2.2-1.2}{2.4-0.4}\\
&=\frac{1}{2}.\\
\end{align*}

This means that an increase of [latex]2[/latex] in the [latex]x-[/latex] direction leads to an increase of [latex]1[/latex] in the [latex]y-[/latex]direction.

The gradient can also be negative. Consider Fig. 7.5.

Here we can set [latex]x_1=0[/latex], [latex]x_2=2[/latex], [latex]y_1=3[/latex] and [latex]y_2=0[/latex]. The gradient of the line is then,

\begin{align*}
m& =\frac{\text{rise}}{\text{run}}\\
&=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\
&=\frac{0-3}{2-0}\\
&=-\frac{3}{2}.\\
\end{align*}

This means that an increase of [latex]2[/latex] in the [latex]x-[/latex] direction leads to a decrease of [latex]3[/latex] in the [latex]y-[/latex]direction.

Positive and negative gradients

Lines that increase their [latex]y[/latex] value as [latex]x[/latex] increases have positive gradients. Lines that decrease their [latex]y[/latex] value as [latex]x[/latex] increases have negative gradients as shown in Fig. 7.6

Here are some exercises to try.

Exercises on gradients and intercepts

1. What is the gradient of the straight line through the points [latex]\left(2,3\right)[/latex] and [latex]\left(-1,5\right)[/latex]

 

Answer

 

[latex]-\dfrac{2}{3}[/latex]

 

 

2. What is the [latex]x-[/latex] intercept, [latex]y-[/latex] intercept and gradient of the line shown in Fig. 7.7 ?

 

Answer

[latex]x-[/latex] intercept [latex]= \left(2,0\right)[/latex]

[latex]y-[/latex] intercept [latex]= \left(0,2\right)[/latex]

gradient [latex]= -1[/latex]

Equation of a straight line given two points

Consider the straight line through the points [latex]\left(x_1,y_1\right)[/latex] and [latex]\left(x_2,y_2\right)[/latex] as shown in Fig. 7.8.

Let [latex]\left(x,y\right)[/latex] be a point on the line and let [latex]\overline{AE}[/latex] be the distance from [latex]A[/latex] to [latex]E[/latex], and let [latex]\overline{AB}[/latex] be the distance from [latex]A[/latex] to [latex]B[/latex] and so on.

The gradient of the line segment from [latex]A[/latex] to [latex]E[/latex] is

\begin{align*}
m_{AE} & =\frac{\text{rise}}{\text{run}}\\
&=\frac{\overline{BE}}{\overline{AB}}\\
&=\frac{y-y_1}{x-x_1}.
\end{align*}

The gradient of the line segment from [latex]A[/latex] to [latex]D[/latex] is

\begin{align*}
m_{AD} & =\frac{\text{rise}}{\text{run}}\\
&=\frac{\overline{CD}}{\overline{AC}}\\
&=\frac{y_2-y_1}{x_2-x_1}.
\end{align*}

Since the gradient of the line is constant, [latex]m_{AE}=m_{AD}[/latex]. That is

\begin{align*}
m_{AE} & =m_{AD}\\
\frac{y-y_1}{x-x_1}&=\frac{y_2-y_1}{x_2-x_1}\\
\end{align*}

Rearranging we get

\begin{align*}
y-y_1&=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\\
y-y_1&=m\left(x-x_1\right)\qquad \left(7.1\right)
\end{align*}

where

\begin{align*}
m&=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}.
\end{align*}

Equation [latex]\left(7.1\right)[/latex] is the equation of a straight line given two points.

To use eqn.[latex]\left(7.1\right)[/latex] we first calculate the gradient and substitute the coordinates of the point [latex]\left(x_1,y_1\right)[/latex].

Example: Finding the equation of a line given two points

1. Find the equation of the line through the points [latex]\left(-1.5,2.5\right)[/latex] and [latex]\left(6,0\right)[/latex].

Solution:

First calculate the gradient [latex]m[/latex]. Set [latex]x_1=-1.5, y_1=2.5, x_2=6[/latex] and [latex]y_2=0[/latex]. Then

\begin{align*}
m&=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\
&=\frac{0-2.5}{6-\left(-1.5 \right)}\\
&=-\frac{2.5}{7.5}\\
&=-\frac{1}{3}.
\end{align*}

Now using eqn.[latex]\left(7.1\right)[/latex],

\begin{align*}
y-y_1&=m\left(x-x_1\right)\\
y-2.5&=-\frac{1}{3}\left(x-\left(-1.5\right)\right)\\
y&=-\frac{1}{3}x+2.
\end{align*}

It is also possible to multiply this equation by 3 to get

\begin{align*}
3y&=-x+6.
\end{align*}

A rearrangement gives

\begin{align*}
3y+x&=6.
\end{align*}

Hence the line through the two given points is [latex]3y+x=6[/latex]  however, the other forms [latex]3y=-x+6[/latex] and [latex]y=-\frac{1}{3}x+2[/latex] are also valid.

 

2. Sketch the graph of the line [latex]3y+2x=4[/latex].

Solution:

First find two points on the graph. Often the [latex]x[/latex] and [latex]y-[/latex] intercepts are convenient but any two points will do.

For the [latex]x-[/latex] intercept, set [latex]y=0[/latex] then the equation becomes [latex]2x=4[/latex]. Hence the [latex]x-[/latex] intercept is [latex]\left(2,0\right)[/latex].

For the [latex]y-[/latex] intercept, set [latex]x=0[/latex] then the equation becomes [latex]3y=4[/latex]. Hence the [latex]y-[/latex] intercept is [latex]\left(0,4/3\right)[/latex].

To sketch the graph, we plot the two points and rule a line through them as shown in Fig. 7.9.

 

Here are some exercises to try.

Equation of a line given the gradient and a point on the line

Equation [latex]\left(7.1\right)[/latex] may also be used to find the equation of a line given a point [latex]\left(x_1,y_1\right)[/latex] and the gradient [latex]m[/latex].

Equation of a line given the gradient and Y- intercept

Rearranging eqn.[latex]\left(7.1\right)[/latex] we have

\begin{align*}
y&=m\left(x-x_1\right)+y_1\\
&=mx-mx_1+y_1\\
&=mx+c \qquad \left(7.2\right)
\end{align*}

where [latex]c=-mx_1+y_1.[/latex]

Note that [latex]c[/latex] may be calculated as we know [latex]m[/latex], [latex]x_1[/latex] and [latex]x_2[/latex]. Also note that when [latex]x=0[/latex], eqn.[latex]\left(7.2\right)[/latex] becomes

\begin{align*}
y&=c
\end{align*}

and so [latex]c[/latex] is the [latex]y-[/latex] intercept of the line.

Equation [latex]\left(7.2\right)[/latex] is the equation of a straight line given the gradient [latex]m[/latex] and the [latex]y-[/latex] intercept [latex]c[/latex].

Click here for an interactive straight line graph. Use the sliders to vary the gradient and y- intercept.

Examples: Equation of a line given the gradient and Y- intercept

1.  A line passes through the point [latex]\left(0,5\right)[/latex] with a gradient of [latex]1/2[/latex]. What is the equation of the line?

Solution:

Let the gradient [latex]m=1/2[/latex] and let [latex]c[/latex] be the [latex]y-[/latex] intercept. Then [latex]c=5[/latex] and, using eqn.[latex]\left(7.2\right)[/latex], we have

\begin{align*}
y&=mx+c\\
&=\frac{1}{2}x+5.
\end{align*}

Hence the equation of the line is [latex]y=\frac{1}{2}x+5[/latex]. Alternate forms of the equation are [latex]2y=x+10[/latex] and [latex]2y-x=10[/latex].

2. What is the equation of the line shown in Fig. 7.10?

Solution:

The first step is to find the [latex]y-[/latex] intercept [latex]c[/latex]. By inspection, the [latex]y-[/latex] intercept is at [latex]\left(0,0.5\right)[/latex]. Hence the [latex]y-[/latex] intercept [latex]c=0.5[/latex] and we set [latex]x_1=0[/latex] and [latex]y_1=0.5[/latex].

Now determine the gradient. We need another point on the line to do this. This can be any point on the graph.  By inspection, the graph goes through the point [latex]\left(2,3.5\right)[/latex] so let [latex]x_2=2[/latex] and [latex]y_2=3.5[/latex] . Then the gradient [latex]m[/latex] is

\begin{align*}
m& =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\
& =\frac{3.5-0.5}{2-0}\\
& =\frac{3}{2}.\\
\end{align*}

Substituting into eqn.[latex]\left(7.2\right)[/latex] gives

\begin{align*}
y&=mx+c\\
&=\frac{3}{2}x+0.5
\end{align*}

Hence the equation is [latex]y=\dfrac{3}{2}x+0.5[/latex]. Alternate forms of the equation are [latex]2y=3x+1[/latex] or [latex]2y-3x=1[/latex].

3. What are the gradient and [latex]y-[/latex] intercept of the line with equation [latex]y=7x+2[/latex]?

Solution:

The line is in the form [latex]y=mx+c[/latex]. Hence [latex]m=7[/latex] and [latex]c=2[/latex]. That is, the gradient of the line is [latex]7[/latex] and the [latex]y-[/latex] intercept is [latex]2[/latex] or [latex]\left(0,2\right)[/latex].

4. What is the gradient and [latex]y-[/latex] intercept of the line with equation [latex]3x+6y=8[/latex]?

Solution:

We need to get the line into the form [latex]y=mx+c[/latex] first. Rearranging,

\begin{align*}
6y& =-3x+8.\\
\end{align*}

Dividing both sides by [latex]6[/latex], we have

\begin{align*}
y& =-\frac{3}{6}x+\frac{8}{6}\\
&=-\frac{1}{2}x+\frac{4}{3}.\\
\end{align*}

Hence the gradient is [latex]1/2[/latex] and the [latex]y-[/latex]intercept is [latex]4/3[/latex] or [latex]\left(0,4/3\right)[/latex].

Here are some exercises to try.

Equations for horizontal and vertical lines

Since a line may be represented in the form

\begin{align*}
y& =mx+c\\
\end{align*}

we see that if the line is horizontal, its gradient [latex]m=0[/latex]. Hence horizontal lines have the equation

\begin{align*}
y& =c\\
\end{align*}

where [latex]c[/latex] is a real number and [latex]\left(0,c\right)[/latex] is the [latex]y-[/latex] intercept.

Vertical lines are not the graph of a linear function because a given value of [latex]x[/latex] has an infinite number of [latex]y[/latex] values. Hence a vertical line is a relation and has the equation [latex]x=a[/latex] where [latex]a[/latex] is a real number. See Figure 7.11.

Examples: Horizontal and vertical lines

1. What is the equation of the vertical line through the point [latex]\left(2,3\right)[/latex]?

Solution:

The graph of the line is shown in Fig.(7.12)

 

Every point on the line has [latex]x-[/latex] co-ordinate [latex]2[/latex]. Hence the equation is [latex]x=2[/latex].

2. What is the equation of the horizontal line through the point [latex]\left(1,-3\right)[/latex]?

Solution:

The graph of the line is shown in Fig.(7.13)

 

Every point on the line has [latex]y-[/latex] co-ordinate [latex]-3[/latex]. Hence the equation is [latex]y=-3[/latex].

Here are some exercises to try.