The Equation of a Straight Line Revision Notes for VCE SSCE Mathematical Methods

The Equation of a Straight Line

Introduction

When working with straight lines in coordinate geometry, we need to determine their equations. To find the equation of a straight line, you generally need two independent pieces of information. This note covers different methods based on what information is given:

Gradient and $y$ -axis intercept
Gradient and a point on the line
Two points on the line

Each method leads to a specific form of the linear equation.

infoNote

The key to finding any straight line equation is having sufficient information to determine both the line's slope and position. Different forms of the equation are suited to different types of given information.

Gradient-intercept form

The gradient-intercept form is one of the most commonly used ways to express the equation of a straight line. Let's explore how this form works.

Understanding the form

Consider a line with gradient $2$ and $y$ -axis intercept $4$ . This line passes through point $A(0, 4)$ on the $y$ -axis. If we take any other point $B(x, y)$ on this line, we can use the gradient formula to establish a relationship.

The gradient of line segment $AB$ is:

\text{Gradient} = \frac{y - 4}{x - 0} = \frac{y - 4}{x}

Since we know the gradient equals $2$ , we can write:

\frac{y - 4}{x} = 2

Rearranging this equation:

y - 4 = 2x

y = 2x + 4

This shows that any point $(x, y)$ on the line satisfies the equation $y = 2x + 4$ .

Conversely, if a point $B(x, y)$ satisfies $y = 2x + 4$ , then $\frac{y - 4}{x} = 2$ , which means the gradient of the line through $A(0, 4)$ and $B(x, y)$ is $2$ , confirming the point lies on our line.

The general gradient-intercept form

Using the same reasoning as above, we can establish the general result:

chatImportant

The line with gradient $m$ and $y$ -axis intercept $c$ has equation:

y = mx + c

Conversely, the line with equation $y = mx + c$ has gradient $m$ and $y$ -axis intercept $c$ .

This is called the gradient-intercept form.

Worked examples with gradient-intercept form

lightbulbExample

Worked Example: Finding gradient and intercept from an equation

Find the gradient and $y$ -axis intercept of the line $y = 3x - 4$ .

Solution:

The equation is already in the form $y = mx + c$ , where m = 3 and c = -4.

Therefore, the gradient is 3 and the $y$ -axis intercept is -4.

lightbulbExample

Worked Example: Writing an equation given gradient and intercept

Find the equation of the line with gradient $-3$ and $y$ -axis intercept $5$ .

Solution:

Using the form $y = mx + c$ with $m = -3$ and $c = 5$ :

y = -3x + 5

lightbulbExample

Worked Example: Rearranging to find gradient and intercept

State the gradient and $y$ -axis intercept of the line $3y + 6x = 9$ .

Solution:

First, rearrange the equation into gradient-intercept form:

3y + 6x = 9

3y = 9 - 6x

y = \frac{9 - 6x}{3}

y = 3 - 2x

y = -2x + 3

Now we can read the gradient and intercept directly: m = -2 and c = 3.

Therefore, the gradient is -2 and the $y$ -axis intercept is 3.

Point-gradient form

Sometimes you know a point on the line and its gradient, but not the $y$ -axis intercept. The point-gradient form is ideal for this situation.

Deriving the point-gradient form

If $A(x_1, y_1)$ is a known point on a line with gradient $m$ , and $P(x, y)$ is any other point on the line, then:

\frac{y - y_1}{x - x_1} = m

Rearranging this equation gives us:

y - y_1 = m(x - x_1)

chatImportant

The point-gradient form of the equation of a straight line is:

y - y_1 = m(x - x_1)

where $(x_1, y_1)$ is a point on the line and $m$ is the gradient.

Worked examples with point-gradient form

lightbulbExample

Worked Example: Using point-gradient form

Find the equation of the line which passes through the point $(-1, 3)$ and has gradient $4$ .

Solution:

Method 1 (using point-gradient form directly):

We have $(x_1, y_1) = (-1, 3)$ and $m = 4$ .

Using the equation $y - y_1 = m(x - x_1)$ :

y - 3 = 4(x - (-1))

y - 3 = 4(x + 1)

y - 3 = 4x + 4

y = 4x + 7

Method 2 (using gradient-intercept form):

Since $m = 4$ , the equation has the form $y = 4x + c$ .

The point $(-1, 3)$ lies on the line, so when $x = -1$ , $y = 3$ :

3 = 4 \times (-1) + c

3 = -4 + c

c = 7

Therefore, the equation is y = 4x + 7.

infoNote

Both methods give the same result. The point-gradient form is often quicker when you don't know the $y$ -intercept, while the gradient-intercept form can be useful if you prefer working with that format.

lightbulbExample

Worked Example: Line with negative gradient

Find the equation of the line that passes through the point $(3, 2)$ and has a gradient of $-2$ .

Solution:

Using the point-gradient form with $(x_1, y_1) = (3, 2)$ and $m = -2$ :

y - 2 = -2(x - 3)

y - 2 = -2x + 6

y = -2x + 8

The equation can also be expressed as 2x + y - 8 = 0 in general form.

A line through two points

When you know two points on a line but not the gradient directly, you can still find the equation by first calculating the gradient.

Method for two points

To find the equation of the line through points $(x_1, y_1)$ and $(x_2, y_2)$ :

Step 1: Calculate the gradient using:

m = \frac{y_2 - y_1}{x_2 - x_1}

Step 2: Use the point-gradient form with either point:

y - y_1 = m(x - x_1)

infoNote

Alternatively, you can find the equation directly by noting that for any point $P(x, y)$ on the line:

\frac{y - y_1}{x - x_1} = m

This combines both steps into one equation.

Worked examples with two points

lightbulbExample

Worked Example: Standard two-point problem

Find the equation of the straight line passing through the points $(1, -2)$ and $(3, 2)$ .

Solution:

First, find the gradient:

m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-2)}{3 - 1} = \frac{4}{2} = 2

Now use $\frac{y - y_1}{x - x_1} = m$ with the point $(x_1, y_1) = (1, -2)$ :

\frac{y - (-2)}{x - 1} = 2

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

y + 2 = 2(x - 1)

y + 2 = 2x - 2

y = 2x - 4

lightbulbExample

Worked Example: Using y-intercept as one point

Find the equation of the straight line with $y$ -axis intercept $-3$ which passes through the point $(1, 10)$ .

Solution:

The $y$ -axis intercept tells us the line passes through (0, -3).

Using $(x_1, y_1) = (0, -3)$ and $(x_2, y_2) = (1, 10)$ :

m = \frac{10 - (-3)}{1 - 0} = \frac{13}{1} = 13

Since we know the $y$ -intercept is $-3$ , we can write directly:

y = 13x - 3

Alternatively, we recognise that a line with $y$ -axis intercept $-3$ has the form $y = mx - 3$ , and we just need to find $m$ using the given point.

Two intercepts (intercept form)

A special case occurs when you know both the x-axis intercept and the y-axis intercept. This leads to a particularly useful form of the equation.

Deriving the intercept form

Consider a line passing through $(a, 0)$ and $(0, b)$ , where $a, b \neq 0$ .

The gradient is:

m = \frac{b - 0}{0 - a} = -\frac{b}{a}

Using the point-gradient form with the point $(a, 0)$ :

y - 0 = -\frac{b}{a}(x - a)

Multiplying both sides by $a$ :

ay = -b(x - a)

ay = -bx + ab

ay + bx = ab

Dividing both sides by $ab$ :

\frac{x}{a} + \frac{y}{b} = 1

chatImportant

The intercept form of the equation of a straight line is:

\frac{x}{a} + \frac{y}{b} = 1

where $a$ is the $x$ -axis intercept and $b$ is the $y$ -axis intercept.

Worked example with intercept form

lightbulbExample

Worked Example: Using the intercept form

Find the equation of the line shown in the graph.

Solution:

From the graph, we can see that point $A$ is at $(0, 4)$ and point $B$ is at $(2, 0)$ .

Therefore, the x-axis intercept is a = 2 and the y-axis intercept is b = 4.

Using the intercept form:

\frac{x}{2} + \frac{y}{4} = 1

Multiplying both sides by $4$ :

2x + y = 4

Rearranging to gradient-intercept form:

y = -2x + 4

Vertical and horizontal lines

These are special cases where the gradient is either zero or undefined.

Horizontal lines

A horizontal line has gradient $m = 0$ . Its equation is simply:

y = c

where $c$ is the $y$ -axis intercept. Every point on the line has the same y-coordinate.

Vertical lines

A vertical line has an undefined gradient. Its equation is:

x = a

where $a$ is the $x$ -axis intercept. Every point on the line has the same x-coordinate.

chatImportant

The equation of a vertical line cannot be written in the form $y = mx + c$ because its gradient is undefined. This is a common source of confusion - remember that vertical lines are a special case requiring the form $x = a$ .

General form of the equation of a straight line

We've seen that all straight lines can be represented by different forms of equations. These can all be rearranged into a common general form.

infoNote

The general form of the equation of a straight line is:

mx + ny + p = 0

where $m$ and $n$ are not both zero.

Conversely, any linear equation of the form $mx + ny + p = 0$ represents a straight line.

Remember!

bookmarkSummary

Key Points to Remember:

Gradient-intercept form ( $y = mx + c$ ): Use when you know the gradient $m$ and $y$ -intercept $c$ . This is the most common form.
Point-gradient form ( $y - y_1 = m(x - x_1)$ ): Use when you know a point $(x_1, y_1)$ on the line and the gradient $m$ .
Two points: First calculate the gradient using $m = \frac{y_2 - y_1}{x_2 - x_1}$ , then use the point-gradient form.
Intercept form ( $\frac{x}{a} + \frac{y}{b} = 1$ ): Use when you know both axis intercepts, where $a$ is the $x$ -intercept and $b$ is the $y$ -intercept.
Special cases: Horizontal lines have equation $y = c$ (gradient = 0), and vertical lines have equation $x = a$ (undefined gradient).

The Equation of a Straight Line (VCE SSCE Mathematical Methods): Revision Notes