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5 Linear relationships
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Stage 5.1-3

5.01 Gradient-intercept form

Lesson
Worksheet
Practice
Lesson

Introduction

Let's have a quick recap of what we have learnt about straight lines on the xy-plane so far.

  • They have a gradient (slope) which is a measure of how steep the line is.

  • They can be increasing (positive gradient) or decreasing (negative gradient).

  • They can be horizontal (zero gradient).

  • They can be vertical (gradient is undefined).

  • They have x-intercepts, y-intercepts or both an x and a y-intercept.

  • The gradient can be calculated using m=\dfrac{\text{rise}}{\text{run}} or m=\dfrac{y_2-y_1}{x_2-x_1}.

  • They have an equation of the form y=mx+c.

Gradient-intercept and general form

The values of m and c have specific meanings.

Exploration

Explore for yourself what these values do by exploring on this interactive. Move the sliders and notice how m affects the gradient and c affects the y-intercept of the line.

Loading interactive...

We can see that the value of m affects the gradient, and that the value of c affects the y-intercept.

For the gradient:

  • If m<0, the gradient is negative and the line is decreasing.

  • If m>0, the gradient is positive and the line is increasing.

  • If m=0, the gradient is 0 and the line is horizontal.

  • The larger the magnitude of m the steeper the line.

For the y-intercept:

  • If c is positive then the line cross the y-axis above the origin.

  • If c is negative then the line cross the y-axis below the origin.

An equation of the form y=mx+c is known as the gradient-intercept form of a line, as we can easily identify both the gradient, m, and the value of the y-intercept, c.

Another useful form for the equation of a straight line is the general form. It looks like this:ax+by+c=0

In this form all coefficients a,b and c are integers and a is positive.

Notice that in this form, the y-intercept cannot be seen in the equation. We would have to substitute x=0 to find it.

The advantages of writing an equation in this form can be seen when:

  • there are fractions involved in the equation (y=\dfrac{-5x}{3}-\dfrac{2}{7} for example). Writing it in general form would be a tidier option.

  • we need to find the point of intersection of two straight lines (and one or both equations involve fractions)

We can convert from one form to another by rearranging the equation. Rearranging the equation is just like solving an equation: we carry out inverse operations to move terms from one side to another, or to change the sign from positive to negative.

Examples

Example 1

A line has the equation 3x-9y-27=0.

a

Express the equation of the line in the form y=mx+c.

Worked Solution
Create a strategy

Use inverse operations to make y the subject.

Apply the idea
\displaystyle 3x-9y-27\displaystyle =\displaystyle 0Write the equation
\displaystyle 3x-9y-27+9y\displaystyle =\displaystyle 0+9yAdd 9y on both sides
\displaystyle 3x-27\displaystyle =\displaystyle 9ySimplify
\displaystyle \dfrac{3x}{9}-\dfrac{27}{9}\displaystyle =\displaystyle \dfrac{9y}{9}Divide each term by 9
\displaystyle y\displaystyle =\displaystyle \dfrac{x}{3}-3Simplify the fractions
b

What is the gradient of the line?

Worked Solution
Create a strategy

Find the coefficient of x from the equation of the form y=mx+c.

Apply the idea

The equation from part (a) was y=\dfrac{x}{3}-3. In this equation the coefficient of x is \dfrac{1}{3}.

So the gradient is m=\dfrac{1}{3}

c

What is the y-intercept of the line?

Worked Solution
Create a strategy

Find the constant term from the equation of the form y=mx+c.

Apply the idea

The equation from part (a) was y=\dfrac{x}{3}-3. In this equation the constant term is -3.

So the y-intercept is c=-3 which has coordinates of (0,-3).

Idea summary

The gradient-intercept form of a straight line:

\displaystyle y=mx+c
\bm{m}
is the gradient of the line.
\bm{c}
is the y-intercept.

The general form of a straight line:

\displaystyle ax+by+c=0
\bm{a}
is a positive integer
\bm{b}
is an integer
\bm{c}
is an integer

The equation from a line

Sometimes we are given the graph of a line, and we are asked to find the equation of the line.

The first thing we want to do is find the  gradient of the line  , which we can do by using any two points (usually the intercepts). Using the coordinates of the two points, we can either use the gradient formula, or by calculating the rise and the run.

We can then identify the y-intercept, by looking where the line crosses the y-axis. Once we have identified these two features, we can write the equation of the line in the form y=mx+c.

Examples

Example 2

Consider the line shown on the coordinate-plane:

-4
-3
-2
-1
1
2
3
4
x
-2
-1
1
2
3
4
5
6
7
y
a

Complete the table of values.

x-1012
y
Worked Solution
Create a strategy

Use the x-values given in the table and find the corresponding points on the line graph.

Apply the idea
-4
-3
-2
-1
1
2
3
4
x
-2
-1
1
2
3
4
5
6
7
y

The coordinates of the points are (-1,7), (0,4), (1,1), and (2,-2).

x-1012
y741-2
b

Linear relations can be written in the form y=mx+c. For this relationship, state the values of m and c.

Worked Solution
Create a strategy

Use the values from the table in part (a) where:

  • m is equal to the change in the y-values for every increase in the x-value by 1, and

  • c is the value of y when x=0.

Apply the idea
x-1012
y741-2

We can see that the y-values are decreasing by 3 for every increase in x-value by 1. So m=-3.

The value of y is 4 when x=0. So c=4.

c

Write the linear equation expressing the relationship between x and y.

Worked Solution
Create a strategy

Substitute the values found in part (b) into the equation y=mx+c.

Apply the idea

y=-3x+4

d

Find the coordinates for the point on the line where x=20.

Worked Solution
Create a strategy

Substitute the given value of x into the equation found in part (c).

Apply the idea
\displaystyle y\displaystyle =\displaystyle -3\times(20)+4Substitute x=20
\displaystyle =\displaystyle -60+4Evaluate the multiplication
\displaystyle =\displaystyle -56Evaluate

The coordinates are (20, -56).

Example 3

Consider the line shown on the coordinate-plane:

-2
-1
1
2
x
-2
-1
1
2
y
a

State the value of the y-intercept.

Worked Solution
Create a strategy

The y-intercept is the point where the line intersects the y-axis, when x=0.

Apply the idea

The line intersects at (0,-1) in the y-axis, so the y-intercept is y=-1.

b

By how much does the y-value change as the x-value increases by 1?

Worked Solution
Create a strategy

Move 1 unit to the right from the y-intercept, then count spaces need to move up to the next point on the line.

Apply the idea
-2
-1
1
2
x
-2
-1
1
2
y

From the y-intercept moving 1 unit to the right and \dfrac12 unit up gets us to the next point on the line.

So, the increase of y-value is \dfrac12.

c

Write the linear equation expressing the relationship between x and y.

Worked Solution
Create a strategy

Use the values found in part (a) and (b) to substitute to the equation y=mx+c, where:

  • m is the increase in y-values when x-value increases by 1, and

  • c is the y-intercept.

Apply the idea

We found from part (b) that the increase in the y-values or m is \dfrac12.

In part (a), we found that the y-intercept or c is -1.

So, the linear equation is y=\dfrac12 x -1 or y=\dfrac{x}{2}-1.

Idea summary

We can find the equation of a line of the form y=mx+c by finding the gradient for the value of m, and the point of y-intercept for the value of c.

Outcomes

MA5.3-8NA

uses formulas to find midpoint, gradient and distance on the Cartesian plane, and applies standard forms of the equation of a straight line

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