1.6.1 Equation of a straight line

What you'll learn

How to calculate and interpret the gradient of a straight line.
How to find equations using $y = m x + c$ , $y-y_1=m(x-x_1)$ and $a x + b y + c = 0$ .
How gradients identify parallel and perpendicular lines.
How straight lines can model relationships in real situations.

Coordinates and straight lines

A point in the coordinate plane is written as $(x, y)$ . The first coordinate gives its horizontal position, and the second gives its vertical position.

A straight line is a set of points following a constant direction. Its steepness does not change, so it has a constant gradient.

Gradient

Definition

Gradient

The gradient, usually written as $m$ , measures the steepness and direction of a straight line. It is the change in $y$ divided by the corresponding change in $x$ :

x.m=\frac{\text{change in }y}{\text{change in }x}.

For two distinct points $x_1,y_1)$ and $x_2,y_2)$ on a non-vertical line,

m=y2−y1x2−x1.m=\frac{y_2-y_1}{x_2-x_1}.

A positive gradient means the line rises from left to right. A negative gradient means it falls from left to right. A horizontal line has gradient zero.

The gradient can be visualised as the vertical rise divided by the horizontal run.

A straight line through two points with a gradient triangle showing rise over run

Example

Finding a gradient from two points

Find the gradient of the line through $A (- 2, 5)$ and $B (4, - 7)$ .

Use the gradient formula with the coordinates in the same order:
$m=−7−54−(−2).m=\frac{-7-5}{4-(-2)}.$
Simplify the changes in the coordinates:
$m=−126=−2.m=\frac{-12}{6}=-2.$
The gradient is negative, so the line should fall from left to right. This agrees with the coordinates: as $x$ increases from $- 2$ to 4, $y$ decreases from 5 to $- 7$ .

Common Mistake

Mixing the coordinate order

If you calculate $y_2-y_1$ in the numerator, you must calculate $x_2-x_1$ in the denominator. Reversing only one subtraction changes the sign incorrectly.

Common Mistake

Vertical lines

For a vertical line, $x_2-x_1=0$ , so the gradient formula would require division by zero. Its gradient is therefore undefined, and its equation has the form $x = k$ , where $k$ is a constant.

The form $y = m x + c$

Definition

Gradient-intercept form

The equation

y = m x + c

is called the gradient-intercept form. Here, $m$ is the gradient and $c$ is the $y$ -intercept, the value of $y$ where the line crosses the $y$ -axis.

At the $y$ -axis, $x = 0$ , so substituting into $y = m x + c$ gives $y = c$ . This explains why the point $(0, c)$ lies on the line.

Example

Finding an equation from a gradient and a point

Find the equation of the line with gradient 3 that passes through $(2, - 1)$ .

Begin with $y = m x + c$ and substitute the known gradient:
$y = 3 x + c .$
The point $(2, - 1)$ lies on the line, so substitute $x = 2$ and $y = - 1$ :
$- 1 = 3 (2) + c .$
Solve for the intercept:
$c = - 7.$
Therefore, the equation is
$y = 3 x - 7.$

Tip

Checking a line equation

Substitute the coordinates of the given point into your final equation. If both sides are equal, the point lies on the line.

The point-gradient form

If you know a gradient and one point, you can avoid finding the intercept first.

Definition

Point-gradient form

A line with gradient $m$ passing through $x_1,y_1)$ has equation

y-y_1=m(x-x_1).

This is the point-gradient form of a straight-line equation.

The formula follows directly from the definition of gradient:

m=y−y1x−x1⇒y−y1=m(x−x1).m=\frac{y-y_1}{x-x_1} \quad\Rightarrow\quad y-y_1=m(x-x_1).

Example

Finding an equation from two points

Find the equation of the line through $(3, 8)$ and $(- 1, - 4)$ .

Calculate the gradient:
$m=8−(−4)3−(−1)=124=3.m=\frac{8-(-4)}{3-(-1)}=\frac{12}{4}=3.$
Use either point in the point-gradient form. Using $(3, 8)$ gives
$y - 8 = 3 (x - 3) .$
Expand and rearrange if required:
$y−8=3x−9y=3x−1.\begin{aligned} y-8&=3x-9\\ y&=3x-1. \end{aligned}$

Key Idea

A gradient and one point determine a line

Once you know the direction of a non-vertical line and one point on it, its position is fixed. This is why a gradient and a point are enough to form its equation.

The general form

Definition

General form

A straight-line equation can be written in the general form

a x + b y + c = 0,

where $a$ , $b$ and $c$ are constants, and $a$ and $b$ are not both zero.

This form includes vertical lines, which cannot be written as $y = m x + c$ . If $b≠0b\neq0$ , rearranging gives

y=−abx−cb,y=-\frac{a}{b}x-\frac{c}{b},

so the gradient is

m=−ab.m=-\frac{a}{b}.

Example

Rearranging a line into general form

Write $y=−23x+5y=-\frac{2}{3}x+5$ in the form $a x + b y + c = 0$ using integer coefficients.

Multiply every term by 3 to remove the fraction:
$3 y = - 2 x + 15.$
Move all terms to one side:
$2 x + 3 y - 15 = 0.$
The coefficients are integers with no common factor, so this is a suitable final form.

Parallel lines

Definition

Parallel lines

Two distinct lines are parallel if they always remain the same distance apart and never meet.

Non-vertical parallel lines have equal gradients. If their gradients are $m_1$ and $m_2$ , then

m_1=m_2.

Vertical lines are also parallel to one another.

Example

Finding a parallel line

Find the equation of the line parallel to $2 x - 5 y + 4 = 0$ and passing through $(3, 1)$ .

Rearrange the given line to identify its gradient:
$−5y=−2x−4y=25x+45.\begin{aligned} -5y&=-2x-4\\ y&=\frac{2}{5}x+\frac{4}{5}. \end{aligned}$
Its gradient is $25\frac{2}{5}$ .
A parallel line has the same gradient, so use the point-gradient form:
$y−1=25(x−3).y-1=\frac{2}{5}(x-3).$
Multiply by 5 and rearrange:
$5y−5=2x−62x−5y−1=0.\begin{aligned} 5y-5&=2x-6\\ 2x-5y-1&=0. \end{aligned}$

Perpendicular lines

Definition

Perpendicular lines

Two lines are perpendicular if they meet at a right angle.

For two non-vertical lines with gradients $m_1$ and $m_2$ , the perpendicular-gradient condition is

m_1m_2=-1.

Therefore, if one gradient is $m$ , the perpendicular gradient is $−1m-\frac{1}{m}$ . This is sometimes described as taking the negative reciprocal.

Example

Finding a perpendicular line

Find the equation of the line perpendicular to $y = 4 x + 7$ and passing through $(8, 2)$ .

The given line has gradient 4, so the perpendicular gradient is
$−14.-\frac{1}{4}.$
Apply the point-gradient form using $(8, 2)$ :
$y−2=−14(x−8).y-2=-\frac{1}{4}(x-8).$
Simplify:
$y−2=−14x+2y=−14x.\begin{aligned} y-2&=-\frac{1}{4}x+2\\ y&=-\frac{1}{4}x. \end{aligned}$

Common Mistake

Changing only the sign

The perpendicular gradient to 4 is not $- 4$ . You must change the sign and take the reciprocal, giving $−14-\frac{1}{4}$ .

Straight-line models

A mathematical model uses mathematics to represent a real situation. A straight-line model has the form

y = m x + c .

In context, $m$ represents the rate at which one quantity changes with another, while $c$ represents the predicted value of $y$ when $x = 0$ .

Example

Modelling a taxi fare

A taxi company charges a fixed booking fee plus a constant amount per mile. A 4-mile journey costs £11, while a 10-mile journey costs £23. Form a model for the cost $C$ pounds of a journey of $d$ miles.

Treat the data as the points $(4, 11)$ and $(10, 23)$ . The rate per mile is the gradient:
$m=23−1110−4=126=2.m=\frac{23-11}{10-4}=\frac{12}{6}=2.$
Write $C = 2 d + c$ and use $(4, 11)$ to find the fixed fee:
$\quad\Rightarrow\quad c=3.$
The model is
$C = 2 d + 3.$
The gradient represents a charge of £2 per mile, and the intercept represents a £3 booking fee.

Exam technique

In the exam

Identify what information you have: two points, or a point and a gradient.
Calculate gradients with a consistent coordinate order and keep fractions exact.
Use equal gradients for parallel lines and $m_1m_2=-1$ for perpendicular lines.
Rearrange into the form requested and verify that the given point satisfies your equation.
In a model, explain the gradient and intercept using the quantities and units in the question.

Self review

Check yourself

Can you find the equation of the line through $(- 2, 7)$ and $(4, - 5)$ ?
What is the gradient of a line perpendicular to $3 x + 2 y - 8 = 0$ ?
In a model $C = 1.5 t + 20$ , what could the gradient and intercept represent?

What you'll learn

How to calculate and interpret the gradient of a straight line.
How to find equations using $y = m x + c$ , $y-y_1=m(x-x_1)$ and $a x + b y + c = 0$ .
How gradients identify parallel and perpendicular lines.
How straight lines can model relationships in real situations.

Coordinates and straight lines

A point in the coordinate plane is written as $(x, y)$ . The first coordinate gives its horizontal position, and the second gives its vertical position.

A straight line is a set of points following a constant direction. Its steepness does not change, so it has a constant gradient.

Gradient

Definition

Gradient

The gradient, usually written as $m$ , measures the steepness and direction of a straight line. It is the change in $y$ divided by the corresponding change in $x$ :

x.m=\frac{\text{change in }y}{\text{change in }x}.

For two distinct points $x_1,y_1)$ and $x_2,y_2)$ on a non-vertical line,

m=y2−y1x2−x1.m=\frac{y_2-y_1}{x_2-x_1}.

A positive gradient means the line rises from left to right. A negative gradient means it falls from left to right. A horizontal line has gradient zero.

The gradient can be visualised as the vertical rise divided by the horizontal run.

A straight line through two points with a gradient triangle showing rise over run

Example

Finding a gradient from two points

Find the gradient of the line through $A (- 2, 5)$ and $B (4, - 7)$ .

Use the gradient formula with the coordinates in the same order:
$m=−7−54−(−2).m=\frac{-7-5}{4-(-2)}.$
Simplify the changes in the coordinates:
$m=−126=−2.m=\frac{-12}{6}=-2.$
The gradient is negative, so the line should fall from left to right. This agrees with the coordinates: as $x$ increases from $- 2$ to 4, $y$ decreases from 5 to $- 7$ .

Common Mistake

Mixing the coordinate order

If you calculate $y_2-y_1$ in the numerator, you must calculate $x_2-x_1$ in the denominator. Reversing only one subtraction changes the sign incorrectly.

Common Mistake

Vertical lines

For a vertical line, $x_2-x_1=0$ , so the gradient formula would require division by zero. Its gradient is therefore undefined, and its equation has the form $x = k$ , where $k$ is a constant.

The form $y = m x + c$

Definition

Gradient-intercept form

The equation

y = m x + c

is called the gradient-intercept form. Here, $m$ is the gradient and $c$ is the $y$ -intercept, the value of $y$ where the line crosses the $y$ -axis.

At the $y$ -axis, $x = 0$ , so substituting into $y = m x + c$ gives $y = c$ . This explains why the point $(0, c)$ lies on the line.

Example

Finding an equation from a gradient and a point

Find the equation of the line with gradient 3 that passes through $(2, - 1)$ .

Begin with $y = m x + c$ and substitute the known gradient:
$y = 3 x + c .$
The point $(2, - 1)$ lies on the line, so substitute $x = 2$ and $y = - 1$ :
$- 1 = 3 (2) + c .$
Solve for the intercept:
$c = - 7.$
Therefore, the equation is
$y = 3 x - 7.$

Tip

Checking a line equation

Substitute the coordinates of the given point into your final equation. If both sides are equal, the point lies on the line.

The point-gradient form

If you know a gradient and one point, you can avoid finding the intercept first.

Definition

Point-gradient form

A line with gradient $m$ passing through $x_1,y_1)$ has equation

y-y_1=m(x-x_1).

This is the point-gradient form of a straight-line equation.

The formula follows directly from the definition of gradient:

m=y−y1x−x1⇒y−y1=m(x−x1).m=\frac{y-y_1}{x-x_1} \quad\Rightarrow\quad y-y_1=m(x-x_1).

Example

Finding an equation from two points

Find the equation of the line through $(3, 8)$ and $(- 1, - 4)$ .

Calculate the gradient:
$m=8−(−4)3−(−1)=124=3.m=\frac{8-(-4)}{3-(-1)}=\frac{12}{4}=3.$
Use either point in the point-gradient form. Using $(3, 8)$ gives
$y - 8 = 3 (x - 3) .$
Expand and rearrange if required:
$y−8=3x−9y=3x−1.\begin{aligned} y-8&=3x-9\\ y&=3x-1. \end{aligned}$

Key Idea

A gradient and one point determine a line

Once you know the direction of a non-vertical line and one point on it, its position is fixed. This is why a gradient and a point are enough to form its equation.

The general form

Definition

General form

A straight-line equation can be written in the general form

a x + b y + c = 0,

where $a$ , $b$ and $c$ are constants, and $a$ and $b$ are not both zero.

This form includes vertical lines, which cannot be written as $y = m x + c$ . If $b≠0b\neq0$ , rearranging gives

y=−abx−cb,y=-\frac{a}{b}x-\frac{c}{b},

so the gradient is

m=−ab.m=-\frac{a}{b}.

Example

Rearranging a line into general form

Write $y=−23x+5y=-\frac{2}{3}x+5$ in the form $a x + b y + c = 0$ using integer coefficients.

Multiply every term by 3 to remove the fraction:
$3 y = - 2 x + 15.$
Move all terms to one side:
$2 x + 3 y - 15 = 0.$
The coefficients are integers with no common factor, so this is a suitable final form.

Parallel lines

Definition

Parallel lines

Two distinct lines are parallel if they always remain the same distance apart and never meet.

Non-vertical parallel lines have equal gradients. If their gradients are $m_1$ and $m_2$ , then

m_1=m_2.

Vertical lines are also parallel to one another.

Example

Finding a parallel line

Find the equation of the line parallel to $2 x - 5 y + 4 = 0$ and passing through $(3, 1)$ .

Rearrange the given line to identify its gradient:
$−5y=−2x−4y=25x+45.\begin{aligned} -5y&=-2x-4\\ y&=\frac{2}{5}x+\frac{4}{5}. \end{aligned}$
Its gradient is $25\frac{2}{5}$ .
A parallel line has the same gradient, so use the point-gradient form:
$y−1=25(x−3).y-1=\frac{2}{5}(x-3).$
Multiply by 5 and rearrange:
$5y−5=2x−62x−5y−1=0.\begin{aligned} 5y-5&=2x-6\\ 2x-5y-1&=0. \end{aligned}$

Perpendicular lines

Definition

Perpendicular lines

Two lines are perpendicular if they meet at a right angle.

For two non-vertical lines with gradients $m_1$ and $m_2$ , the perpendicular-gradient condition is

m_1m_2=-1.

Therefore, if one gradient is $m$ , the perpendicular gradient is $−1m-\frac{1}{m}$ . This is sometimes described as taking the negative reciprocal.

Example

Finding a perpendicular line

Find the equation of the line perpendicular to $y = 4 x + 7$ and passing through $(8, 2)$ .

The given line has gradient 4, so the perpendicular gradient is
$−14.-\frac{1}{4}.$
Apply the point-gradient form using $(8, 2)$ :
$y−2=−14(x−8).y-2=-\frac{1}{4}(x-8).$
Simplify:
$y−2=−14x+2y=−14x.\begin{aligned} y-2&=-\frac{1}{4}x+2\\ y&=-\frac{1}{4}x. \end{aligned}$

Common Mistake

Changing only the sign

The perpendicular gradient to 4 is not $- 4$ . You must change the sign and take the reciprocal, giving $−14-\frac{1}{4}$ .

Straight-line models

A mathematical model uses mathematics to represent a real situation. A straight-line model has the form

y = m x + c .

In context, $m$ represents the rate at which one quantity changes with another, while $c$ represents the predicted value of $y$ when $x = 0$ .

Example

Modelling a taxi fare

Treat the data as the points $(4, 11)$ and $(10, 23)$ . The rate per mile is the gradient:
$m=23−1110−4=126=2.m=\frac{23-11}{10-4}=\frac{12}{6}=2.$
Write $C = 2 d + c$ and use $(4, 11)$ to find the fixed fee:
$\quad\Rightarrow\quad c=3.$
The model is
$C = 2 d + 3.$
The gradient represents a charge of £2 per mile, and the intercept represents a £3 booking fee.

Exam technique

In the exam

Identify what information you have: two points, or a point and a gradient.
Calculate gradients with a consistent coordinate order and keep fractions exact.
Use equal gradients for parallel lines and $m_1m_2=-1$ for perpendicular lines.
Rearrange into the form requested and verify that the given point satisfies your equation.
In a model, explain the gradient and intercept using the quantities and units in the question.

Self review

Check yourself

Can you find the equation of the line through $(- 2, 7)$ and $(4, - 5)$ ?
What is the gradient of a line perpendicular to $3 x + 2 y - 8 = 0$ ?
In a model $C = 1.5 t + 20$ , what could the gradient and intercept represent?

What you'll learn

Coordinates and straight lines

Gradient

The form $y = m x + c$

The point-gradient form

The general form

Parallel lines

Perpendicular lines

Straight-line models

Review 1.6.1 Equation of a straight line by teaching Genie

1.6.1 Equation of a straight line

What you'll learn

Coordinates and straight lines

Gradient

The form $y = m x + c$

The point-gradient form

The general form

Parallel lines

Perpendicular lines

Straight-line models

Review 1.6.1 Equation of a straight line by teaching Genie

1.6.1 Equation of a straight line

What you'll learn

Coordinates and straight lines

Gradient

The form y=mx+cy=mx+cy=mx+c

The point-gradient form

The general form

Parallel lines

Perpendicular lines

Straight-line models

Review 1.6.1 Equation of a straight line by teaching Genie

1.6.1 Equation of a straight line Revision Guide

1.6.1 Equation of a straight line

What you'll learn

Coordinates and straight lines

Gradient

The form y=mx+cy=mx+cy=mx+c

The point-gradient form

The general form

Parallel lines

Perpendicular lines

Straight-line models

Review 1.6.1 Equation of a straight line by teaching Genie

1.6.1 Equation of a straight line Revision Guide

The form $y = m x + c$

The form $y = m x + c$