Straight Line Graphs

What you'll learn

How to find the gradient of a line from two points or from its equation.
How to write equations of straight lines in forms such as $y = m x + c$ and $a x + b y + c = 0$ .
How to decide whether lines are parallel or perpendicular.
How to use midpoints, distances, intercepts and intersections in coordinate geometry problems.

Coordinates and gradient

A point on a graph is written as an ordered pair, such as $(3, - 2)$ . The first number is the $x$ -coordinate, and the second is the $y$ -coordinate.

The origin is the point $(0, 0)$ , where the axes meet.

Definition

Gradient

The gradient of a straight line measures its steepness. Between two points $x_1, y_1)$ and $x_2, y_2)$ , the gradient is

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

Think of gradient as:

A gradient triangle shows that gradient is the vertical change divided by the horizontal change between two points on a line.

x\text{gradient} = \frac{\text{change in } y}{\text{change in } x}

A straight line through two points with a gradient triangle showing vertical and horizontal change

Common Mistake

Subtracting in different orders

If you do $y_2 - y_1$ on the top, you must do $x_2 - x_1$ on the bottom. Do not swap the order halfway through.

Example

Finding the gradient between two points

A line passes through $(- 1, 6)$ and $(5, - 3)$ . Find its gradient.

Identify the two points:
$(x1,y1)=(−1,6),(x2,y2)=(5,−3)(x_1, y_1) = (-1, 6), \qquad (x_2, y_2) = (5, -3)$
Substitute into the gradient formula:
$\frac{-3 - 6}{5 - (-1)}$
Simplify the numerator and denominator:
$\frac{-9}{6} = -\frac{3}{2}$
The gradient of the line is $−32-\frac{3}{2}$ .

Equations of straight lines

The most common form of a straight-line equation is:

y = m x + c

Here, $m$ is the gradient and $c$ is the $y$ -intercept.

The equation y = mx + c links the gradient of the line to its crossing point on the y-axis.

Definition

The y-intercept

The $y$ -intercept is the point where the line crosses the $y$ -axis. In $y = m x + c$ , the line crosses the $y$ -axis at $(0, c)$ .

You can find the equation of a line if you know:

its gradient and one point on it, or
two points on it.

Key Idea

Equation from gradient and a point

If you know the gradient $m$ and a point $x_1, y_1)$ , substitute into $y = m x + c$ to find $c$ .

Example

Equation of a line through two points

Find an equation of the line through $(2, 7)$ and $(6, - 1)$ .

First find the gradient:
$\frac{-1 - 7}{6 - 2} = \frac{-8}{4} = -2$
Start with $y = m x + c$ . Since the gradient is -2:
$y = - 2 x + c$
Substitute one point, for example $(2, 7)$ :
$7 = - 2 (2) + c$
Solve for $c$ :
$7 = - 4 + c$
Therefore $c = 11$ , so the equation is:
$y = - 2 x + 11$

Rearranging into $a x + b y + c = 0$

In AS questions, you are often asked to give your answer in the form:

a x + b y + c = 0

where $a$ , $b$ and $c$ are integers.

Definition

General form

The form $a x + b y + c = 0$ is called the general form of a straight-line equation. It collects all terms on one side and leaves zero on the other side.

Example

Writing an equation in general form

A line passes through $(- 2, 4)$ and $(4, 1)$ . Find its equation in the form $a x + b y + c = 0$ .

Find the gradient:
$\frac{1 - 4}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2}$
Use $y = m x + c$ :
$-\frac{1}{2}x + c$
Substitute the point $(4, 1)$ :
$-\frac{1}{2}(4) + c$
Solve for $c$ :
$1 = - 2 + c$
So $c = 3$ , giving:
$-\frac{1}{2}x + 3$
Multiply by 2 to remove the fraction:
$2 y = - x + 6$
Rearrange into general form:
$x + 2 y - 6 = 0$

Tip

Removing fractions

If your equation contains fractions, multiply every term by the denominator before rearranging into $a x + b y + c = 0$ .

Intercepts and areas with the axes

A line may cross the coordinate axes at two useful points:

The $x$ -intercept is where $y = 0$ .
The $y$ -intercept is where $x = 0$ .

These are often used to form a triangle with the origin.

The x- and y-intercepts of a line can form a right-angled triangle with the origin.

Example

Finding intercepts and an area

The line $3 x + 2 y - 12 = 0$ meets the axes at points $A$ and $B$ . Find the area of triangle $A O B$ , where $O$ is the origin.

To find the $x$ -intercept, set $y = 0$ :
$3 x + 2 (0) - 12 = 0$
Solve for $x$ :
$3 x = 12$
So $x = 4$ , meaning $A = (4, 0)$ .
To find the $y$ -intercept, set $x = 0$ :
$3 (0) + 2 y - 12 = 0$
Solve for $y$ :
$2 y = 12$
So $y = 6$ , meaning $B = (0, 6)$ .
The triangle has base 4 and height 6, so its area is:
$12×4×6=12\frac{1}{2} \times 4 \times 6 = 12$

Parallel and perpendicular lines

Definition

Parallel lines

Two straight lines are parallel if they have the same gradient and never meet.

Definition

Perpendicular lines

Two straight lines are perpendicular if they meet at 90°. Their gradients multiply to give -1.

Parallel lines have equal gradients, while perpendicular lines meet at a right angle and have negative reciprocal gradients.

m_1m_2 = -1

So if one line has gradient 3, a perpendicular line has gradient $−13-\frac{1}{3}$ .

If one line has gradient $−25-\frac{2}{5}$ , a perpendicular line has gradient $52\frac{5}{2}$ .

Example

Deciding whether two lines are parallel or perpendicular

Line $l_1$ has equation $4 x + 2 y - 10 = 0$ . Line $l_2$ passes through $(1, 5)$ and $(5, - 3)$ . Decide whether the lines are parallel, perpendicular or neither.

Rearrange $l_1$ into $y = m x + c$ form:
$4 x + 2 y - 10 = 0$
Make $y$ the subject:
$2 y = - 4 x + 10$
Divide by 2, so the gradient of $l_1$ is -2:
$y = - 2 x + 5$
Find the gradient of $l_2$ using its two points:
$\frac{-3 - 5}{5 - 1} = \frac{-8}{4} = -2$
Both gradients are the same, so the lines are parallel.

Common Mistake

Perpendicular does not mean opposite signs only

Gradients 2 and -2 are not perpendicular. For perpendicular lines, the gradients must multiply to -1.

Midpoints, distances and perpendicular bisectors

The midpoint of a line segment is the point halfway between its endpoints.

Definition

Midpoint formula

For points $x_1, y_1)$ and $x_2, y_2)$ , the midpoint is

(x1+x22,y1+y22)\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

The distance between two points comes from Pythagoras’ theorem.

Definition

Distance formula

For points $x_1, y_1)$ and $x_2, y_2)$ , the distance between them is

(x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

A perpendicular bisector cuts a line segment exactly in half and meets it at 90°.

A perpendicular bisector passes through the midpoint of a segment and is at 90° to it.

A line segment AB with midpoint M and its perpendicular bisector shown at right angles

Example

Finding a perpendicular bisector

Points $A (- 4, 1)$ and $B (2, 5)$ are joined by a straight line. Find the equation of the perpendicular bisector of $A B$ in the form $a x + b y + c = 0$ .

Find the gradient of $A B$ :
$mAB=5−12−(−4)=46=23m_{AB} = \frac{5 - 1}{2 - (-4)} = \frac{4}{6} = \frac{2}{3}$
The perpendicular gradient is the negative reciprocal:
$-\frac{3}{2}$
Find the midpoint of $A B$ :
$(−4+22,1+52)=(−1,3)\left(\frac{-4 + 2}{2}, \frac{1 + 5}{2}\right) = (-1, 3)$
Use $y = m x + c$ with midpoint $(- 1, 3)$ :
$-\frac{3}{2}(-1) + c$
Solve for $c$ :
$\frac{3}{2} + c$
So $\frac{3}{2}$ , giving:
$-\frac{3}{2}x + \frac{3}{2}$
Multiply by 2 and rearrange:
$2 y = - 3 x + 3$
The perpendicular bisector is:
$3 x + 2 y - 3 = 0$

Intersections of straight lines

When two lines intersect, their equations are both true at the same point. You find the point of intersection by solving the two equations simultaneously.

The solution of two simultaneous linear equations is the coordinate where their graphs intersect.

Example

Finding the point where two lines meet

Find the coordinates where $2 x - y + 1 = 0$ and $x + 3 y - 14 = 0$ intersect.

Rearrange the first equation to make $y$ the subject:
$y = 2 x + 1$
Substitute this into the second equation:
$x + 3 (2 x + 1) - 14 = 0$
Expand and simplify:
$x + 6 x + 3 - 14 = 0$
Solve for $x$ :
$7 x - 11 = 0$
So $\frac{11}{7}$ .
Substitute into $y = 2 x + 1$ :
$2\left(\frac{11}{7}\right) + 1$
Therefore:
$\frac{22}{7} + \frac{7}{7} = \frac{29}{7}$
The intersection point is $(117,297)\left(\frac{11}{7}, \frac{29}{7}\right)$ .

Tip

Check your intersection

Substitute your final coordinates into both original equations. If they work in both, your intersection is correct.

Linear models

A linear model assumes that two quantities are linked by a straight-line relationship. At AS Level, this often means using two data points to form an equation.

For example, if $A$ is an amount and $n$ is the number of years after a starting year, a linear model may look like:

A = mn + c

Here, $m$ is the rate of change per year, and $c$ is the starting value when $n = 0$ .

A linear model represents a constant rate of change from an initial value.

Example

Forming and commenting on a linear model

In 2000, a town had population 40 thousand. In 2010, it had population 46 thousand. Let $P$ be the population in thousands, $n$ years after 2000. Form a linear model, then comment if the actual population in 2020 was 51 thousand.

Write the two points using $n$ and $P$ :
$\qquad (10, 46)$
Find the gradient:
$\frac{46 - 40}{10 - 0} = \frac{6}{10} = 0.6$
Since $P = 40$ when $n = 0$ , the intercept is 40.
The model is:
$P = 0.6 n + 40$
For 2020, $n = 20$ , so the model predicts:
$P = 0.6 (20) + 40 = 52$
The model predicts 52 thousand, but the actual value is 51 thousand. It is close, so the model is fairly suitable, although it slightly overestimates.

Common Mistake

Models are not facts

A straight-line model is an assumption. A comment question usually wants you to compare the model’s prediction with the real value and say whether the difference is large or small in context.

Exam technique

In the exam

Start by finding the gradient whenever a question involves two points, parallel lines or perpendicular lines.
If asked for $a x + b y + c = 0$ , clear fractions and move every term to one side.
For a perpendicular bisector, you need both the midpoint and the perpendicular gradient.
For axis intercepts, set $y = 0$ for the $x$ -intercept and set $x = 0$ for the $y$ -intercept.

Self review

Check yourself

Can you find the equation of a line through two given points?
Can you explain how to tell whether two lines are parallel or perpendicular?
Can you find the perpendicular bisector of a line segment using its midpoint?

What you'll learn

How to find the gradient of a line from two points or from its equation.
How to write equations of straight lines in forms such as $y = m x + c$ and $a x + b y + c = 0$ .
How to decide whether lines are parallel or perpendicular.
How to use midpoints, distances, intercepts and intersections in coordinate geometry problems.

Coordinates and gradient

A point on a graph is written as an ordered pair, such as $(3, - 2)$ . The first number is the $x$ -coordinate, and the second is the $y$ -coordinate.

The origin is the point $(0, 0)$ , where the axes meet.

Definition

Gradient

The gradient of a straight line measures its steepness. Between two points $x_1, y_1)$ and $x_2, y_2)$ , the gradient is

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

Think of gradient as:

A gradient triangle shows that gradient is the vertical change divided by the horizontal change between two points on a line.

x\text{gradient} = \frac{\text{change in } y}{\text{change in } x}

A straight line through two points with a gradient triangle showing vertical and horizontal change

Common Mistake

Subtracting in different orders

If you do $y_2 - y_1$ on the top, you must do $x_2 - x_1$ on the bottom. Do not swap the order halfway through.

Example

Finding the gradient between two points

A line passes through $(- 1, 6)$ and $(5, - 3)$ . Find its gradient.

Identify the two points:
$(x1,y1)=(−1,6),(x2,y2)=(5,−3)(x_1, y_1) = (-1, 6), \qquad (x_2, y_2) = (5, -3)$
Substitute into the gradient formula:
$\frac{-3 - 6}{5 - (-1)}$
Simplify the numerator and denominator:
$\frac{-9}{6} = -\frac{3}{2}$
The gradient of the line is $−32-\frac{3}{2}$ .

Equations of straight lines

The most common form of a straight-line equation is:

y = m x + c

Here, $m$ is the gradient and $c$ is the $y$ -intercept.

The equation y = mx + c links the gradient of the line to its crossing point on the y-axis.

Definition

The y-intercept

The $y$ -intercept is the point where the line crosses the $y$ -axis. In $y = m x + c$ , the line crosses the $y$ -axis at $(0, c)$ .

You can find the equation of a line if you know:

its gradient and one point on it, or
two points on it.

Key Idea

Equation from gradient and a point

If you know the gradient $m$ and a point $x_1, y_1)$ , substitute into $y = m x + c$ to find $c$ .

Example

Equation of a line through two points

Find an equation of the line through $(2, 7)$ and $(6, - 1)$ .

First find the gradient:
$\frac{-1 - 7}{6 - 2} = \frac{-8}{4} = -2$
Start with $y = m x + c$ . Since the gradient is -2:
$y = - 2 x + c$
Substitute one point, for example $(2, 7)$ :
$7 = - 2 (2) + c$
Solve for $c$ :
$7 = - 4 + c$
Therefore $c = 11$ , so the equation is:
$y = - 2 x + 11$

Rearranging into $a x + b y + c = 0$

In AS questions, you are often asked to give your answer in the form:

a x + b y + c = 0

where $a$ , $b$ and $c$ are integers.

Definition

General form

The form $a x + b y + c = 0$ is called the general form of a straight-line equation. It collects all terms on one side and leaves zero on the other side.

Example

Writing an equation in general form

A line passes through $(- 2, 4)$ and $(4, 1)$ . Find its equation in the form $a x + b y + c = 0$ .

Find the gradient:
$\frac{1 - 4}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2}$
Use $y = m x + c$ :
$-\frac{1}{2}x + c$
Substitute the point $(4, 1)$ :
$-\frac{1}{2}(4) + c$
Solve for $c$ :
$1 = - 2 + c$
So $c = 3$ , giving:
$-\frac{1}{2}x + 3$
Multiply by 2 to remove the fraction:
$2 y = - x + 6$
Rearrange into general form:
$x + 2 y - 6 = 0$

Tip

Removing fractions

If your equation contains fractions, multiply every term by the denominator before rearranging into $a x + b y + c = 0$ .

Intercepts and areas with the axes

A line may cross the coordinate axes at two useful points:

The $x$ -intercept is where $y = 0$ .
The $y$ -intercept is where $x = 0$ .

These are often used to form a triangle with the origin.

The x- and y-intercepts of a line can form a right-angled triangle with the origin.

Example

Finding intercepts and an area

The line $3 x + 2 y - 12 = 0$ meets the axes at points $A$ and $B$ . Find the area of triangle $A O B$ , where $O$ is the origin.

To find the $x$ -intercept, set $y = 0$ :
$3 x + 2 (0) - 12 = 0$
Solve for $x$ :
$3 x = 12$
So $x = 4$ , meaning $A = (4, 0)$ .
To find the $y$ -intercept, set $x = 0$ :
$3 (0) + 2 y - 12 = 0$
Solve for $y$ :
$2 y = 12$
So $y = 6$ , meaning $B = (0, 6)$ .
The triangle has base 4 and height 6, so its area is:
$12×4×6=12\frac{1}{2} \times 4 \times 6 = 12$

Parallel and perpendicular lines

Definition

Parallel lines

Two straight lines are parallel if they have the same gradient and never meet.

Definition

Perpendicular lines

Two straight lines are perpendicular if they meet at 90°. Their gradients multiply to give -1.

Parallel lines have equal gradients, while perpendicular lines meet at a right angle and have negative reciprocal gradients.

m_1m_2 = -1

So if one line has gradient 3, a perpendicular line has gradient $−13-\frac{1}{3}$ .

If one line has gradient $−25-\frac{2}{5}$ , a perpendicular line has gradient $52\frac{5}{2}$ .

Example

Deciding whether two lines are parallel or perpendicular

Line $l_1$ has equation $4 x + 2 y - 10 = 0$ . Line $l_2$ passes through $(1, 5)$ and $(5, - 3)$ . Decide whether the lines are parallel, perpendicular or neither.

Rearrange $l_1$ into $y = m x + c$ form:
$4 x + 2 y - 10 = 0$
Make $y$ the subject:
$2 y = - 4 x + 10$
Divide by 2, so the gradient of $l_1$ is -2:
$y = - 2 x + 5$
Find the gradient of $l_2$ using its two points:
$\frac{-3 - 5}{5 - 1} = \frac{-8}{4} = -2$
Both gradients are the same, so the lines are parallel.

Common Mistake

Perpendicular does not mean opposite signs only

Gradients 2 and -2 are not perpendicular. For perpendicular lines, the gradients must multiply to -1.

Midpoints, distances and perpendicular bisectors

The midpoint of a line segment is the point halfway between its endpoints.

Definition

Midpoint formula

For points $x_1, y_1)$ and $x_2, y_2)$ , the midpoint is

(x1+x22,y1+y22)\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

The distance between two points comes from Pythagoras’ theorem.

Definition

Distance formula

For points $x_1, y_1)$ and $x_2, y_2)$ , the distance between them is

(x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

A perpendicular bisector cuts a line segment exactly in half and meets it at 90°.

A perpendicular bisector passes through the midpoint of a segment and is at 90° to it.

A line segment AB with midpoint M and its perpendicular bisector shown at right angles

Example

Finding a perpendicular bisector

Points $A (- 4, 1)$ and $B (2, 5)$ are joined by a straight line. Find the equation of the perpendicular bisector of $A B$ in the form $a x + b y + c = 0$ .

Find the gradient of $A B$ :
$mAB=5−12−(−4)=46=23m_{AB} = \frac{5 - 1}{2 - (-4)} = \frac{4}{6} = \frac{2}{3}$
The perpendicular gradient is the negative reciprocal:
$-\frac{3}{2}$
Find the midpoint of $A B$ :
$(−4+22,1+52)=(−1,3)\left(\frac{-4 + 2}{2}, \frac{1 + 5}{2}\right) = (-1, 3)$
Use $y = m x + c$ with midpoint $(- 1, 3)$ :
$-\frac{3}{2}(-1) + c$
Solve for $c$ :
$\frac{3}{2} + c$
So $\frac{3}{2}$ , giving:
$-\frac{3}{2}x + \frac{3}{2}$
Multiply by 2 and rearrange:
$2 y = - 3 x + 3$
The perpendicular bisector is:
$3 x + 2 y - 3 = 0$

Intersections of straight lines

When two lines intersect, their equations are both true at the same point. You find the point of intersection by solving the two equations simultaneously.

The solution of two simultaneous linear equations is the coordinate where their graphs intersect.

Example

Finding the point where two lines meet

Find the coordinates where $2 x - y + 1 = 0$ and $x + 3 y - 14 = 0$ intersect.

Rearrange the first equation to make $y$ the subject:
$y = 2 x + 1$
Substitute this into the second equation:
$x + 3 (2 x + 1) - 14 = 0$
Expand and simplify:
$x + 6 x + 3 - 14 = 0$
Solve for $x$ :
$7 x - 11 = 0$
So $\frac{11}{7}$ .
Substitute into $y = 2 x + 1$ :
$2\left(\frac{11}{7}\right) + 1$
Therefore:
$\frac{22}{7} + \frac{7}{7} = \frac{29}{7}$
The intersection point is $(117,297)\left(\frac{11}{7}, \frac{29}{7}\right)$ .

Tip

Check your intersection

Substitute your final coordinates into both original equations. If they work in both, your intersection is correct.

Linear models

A linear model assumes that two quantities are linked by a straight-line relationship. At AS Level, this often means using two data points to form an equation.

For example, if $A$ is an amount and $n$ is the number of years after a starting year, a linear model may look like:

A = mn + c

Here, $m$ is the rate of change per year, and $c$ is the starting value when $n = 0$ .

A linear model represents a constant rate of change from an initial value.

Example

Forming and commenting on a linear model

Write the two points using $n$ and $P$ :
$\qquad (10, 46)$
Find the gradient:
$\frac{46 - 40}{10 - 0} = \frac{6}{10} = 0.6$
Since $P = 40$ when $n = 0$ , the intercept is 40.
The model is:
$P = 0.6 n + 40$
For 2020, $n = 20$ , so the model predicts:
$P = 0.6 (20) + 40 = 52$
The model predicts 52 thousand, but the actual value is 51 thousand. It is close, so the model is fairly suitable, although it slightly overestimates.

Common Mistake

Models are not facts

A straight-line model is an assumption. A comment question usually wants you to compare the model’s prediction with the real value and say whether the difference is large or small in context.

Exam technique

In the exam

Start by finding the gradient whenever a question involves two points, parallel lines or perpendicular lines.
If asked for $a x + b y + c = 0$ , clear fractions and move every term to one side.
For a perpendicular bisector, you need both the midpoint and the perpendicular gradient.
For axis intercepts, set $y = 0$ for the $x$ -intercept and set $x = 0$ for the $y$ -intercept.

Self review

Check yourself

Can you find the equation of a line through two given points?
Can you explain how to tell whether two lines are parallel or perpendicular?
Can you find the perpendicular bisector of a line segment using its midpoint?

What you'll learn

Coordinates and gradient

Equations of straight lines

Rearranging into $a x + b y + c = 0$

Intercepts and areas with the axes

Parallel and perpendicular lines

Midpoints, distances and perpendicular bisectors

Intersections of straight lines

Linear models

Recap questions

Review Straight Line Graphs by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Straight Line Graphs

What you'll learn

Coordinates and gradient

Equations of straight lines

Rearranging into $a x + b y + c = 0$

Intercepts and areas with the axes

Parallel and perpendicular lines

Midpoints, distances and perpendicular bisectors

Intersections of straight lines

Linear models

Recap questions

Review Straight Line Graphs by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Straight Line Graphs

What you'll learn

Coordinates and gradient

Equations of straight lines

Rearranging into ax+by+c=0ax + by + c = 0ax+by+c=0

Intercepts and areas with the axes

Parallel and perpendicular lines

Midpoints, distances and perpendicular bisectors

Intersections of straight lines

Linear models

Recap questions

Review Straight Line Graphs by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Straight Line Graphs Revision Guide

Straight Line Graphs

What you'll learn

Coordinates and gradient

Equations of straight lines

Rearranging into ax+by+c=0ax + by + c = 0ax+by+c=0

Intercepts and areas with the axes

Parallel and perpendicular lines

Midpoints, distances and perpendicular bisectors

Intersections of straight lines

Linear models

Recap questions

Review Straight Line Graphs by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Straight Line Graphs Revision Guide

Rearranging into $a x + b y + c = 0$

Rearranging into $a x + b y + c = 0$