Straight-line graphs: gradients and equations

A straight line has a constant rate of change. Its equation connects every point on the line, its gradient describes the line’s direction and steepness, and its intercept records where it crosses an axis.

These ideas are essential for coordinate geometry, simultaneous equations, differentiation, kinematics and mathematical modelling.

Before you begin

You should be able to:

• plot and read coordinates;
• substitute into an expression;
• rearrange and solve linear equations;
• calculate accurately with negative numbers and fractions.

Review coordinate geometry, linear equations or rearranging formulae if needed.

Gradient measures change

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

$$\boxed{m=\frac{\text{change in }y}{\text{change in }x} =\frac{y_2-y_1}{x_2-x_1}}.$$

Gradient is a ratio, not an angle or a length. It tells you how much $y$ changes when $x$ increases by $1$.

Gradient	Behaviour from left to right
$m>0$	the line rises
$m<0$	the line falls
$m=0$	the line is horizontal
undefined	the line is vertical

The steeper line has the greater value of $|m|$. Thus a line of gradient $-5$ is steeper than one of gradient $2$, even though $-5<2$.

Worked example 1: gradient from two points

Find the gradient of the line through $A(-3,5)$ and $B(5,-1)$.

Use the same subtraction order in numerator and denominator:

$$m=\frac{-1-5}{5-(-3)} =\frac{-6}{8} =-\frac34.$$

So every increase of $4$ in $x$ produces a decrease of $3$ in $y$.

Reversing both subtraction orders gives the same result:

$$\frac{5-(-1)}{-3-5}=\frac6{-8}=-\frac34.$$

Common misconception: mismatched subtraction

If you calculate $y_2-y_1$ on top, you must calculate $x_2-x_1$ underneath. Mixing the orders changes the sign incorrectly.

Self-check 1

Find the gradient through $(2,-4)$ and $(-6,8)$.

Answer

$$m=\frac{8-(-4)}{-6-2}=\frac{12}{-8}=-\frac32.$$

The equation $y=mx+c$

Every non-vertical straight line can be written as

$$\boxed{y=mx+c},$$

where:

• $m$ is the gradient;
• $c$ is the $y$ intercept, the value of $y$ when $x=0$.

For example, in

$$y=-2x+7,$$

the gradient is $-2$ and the line crosses the $y$ axis at $(0,7)$.

The equation is a rule for generating points. If $x=3$, then

$$y=-2(3)+7=1,$$

so $(3,1)$ lies on the line.

Worked example 2: draw a line efficiently

Sketch $y=\dfrac23x-2$.

1. The $y$ intercept is $(0,-2)$, so plot that point.
2. The gradient $\dfrac23$ means a change of $3$ right and $2$ up.
3. This reaches $(3,0)$. Repeat to reach $(6,2)$.
4. Join the points with a straight line and extend it in both directions.

A table gives the same points:

$x$	$0$	$3$	$6$
$y$	$-2$	$0$	$2$

Choose convenient $x$ values. When the gradient contains thirds, multiples of $3$ avoid awkward plotting decimals.

Axis intercepts

To find where a graph crosses an axis:

• on the $y$ axis, $x=0$;
• on the $x$ axis, $y=0$.

Worked example 3: find both intercepts

For $3x+2y=12$:

When $x=0$,

$$2y=12\quad\Longrightarrow\quad y=6,$$

so the $y$ intercept is $(0,6)$.

When $y=0$,

$$3x=12\quad\Longrightarrow\quad x=4,$$

so the $x$ intercept is $(4,0)$. Two points determine a unique line, so these intercepts are enough to draw it.

Common misconception: confusing $c$ with the $x$ intercept

In $y=mx+c$, $c$ is the $y$ intercept because setting $x=0$ gives $y=c$. The $x$ intercept must usually be calculated by setting $y=0$.

Recognising the gradient in other forms

An equation may need rearranging before its gradient and $y$ intercept are visible.

Worked example 4: rearrange to $y=mx+c$

Find the gradient and intercepts of

$$4x-2y=10.$$

Make $y$ the subject:

$$\begin{aligned} -2y&=10-4x\\ y&=2x-5. \end{aligned}$$

Therefore the gradient is $2$ and the $y$ intercept is $(0,-5)$.

For the $x$ intercept, set $y=0$ in either form:

$$4x=10\quad\Longrightarrow\quad x=\frac52.$$

Hence the $x$ intercept is $\left(\dfrac52,0\right)$.

Do not read the coefficient $4$ in $4x-2y=10$ as the gradient. The equation must first be rearranged into a form where $y$ is the subject.

Finding the equation of a line

Different information suggests different methods.

Given the gradient and $y$ intercept

Substitute directly into $y=mx+c$.

Worked example 5

A line has gradient $-\dfrac52$ and crosses the $y$ axis at $(0,4)$. Its equation is

$$\boxed{y=-\frac52x+4}.$$

Given a gradient and one point

If a line of gradient $m$ passes through $(x_1,y_1)$, use the point-gradient form

$$\boxed{y-y_1=m(x-x_1)}.$$

This says that the change in $y$ from the known point equals $m$ times the change in $x$.

Worked example 6: point and gradient

Find the equation of the line with gradient $3$ through $(-2,5)$.

Substitute $m=3$, $x_1=-2$ and $y_1=5$:

$$y-5=3(x-(-2))=3(x+2).$$

Rearrange if $y=mx+c$ is required:

$$y-5=3x+6 \quad\Longrightarrow\quad \boxed{y=3x+11}.$$

Check the known point:

$$3(-2)+11=5.$$

An equally valid method is to begin with $y=3x+c$, substitute $(-2,5)$, and solve $5=-6+c$.

Given two points

First calculate the gradient. Then use either point to find the equation.

Worked example 7: equation through two points

Find the equation through $P(2,7)$ and $Q(8,-2)$.

First,

$$m=\frac{-2-7}{8-2}=\frac{-9}{6}=-\frac32.$$

Use $P(2,7)$ in point-gradient form:

$$y-7=-\frac32(x-2).$$

Expand:

$$y-7=-\frac32x+3,$$

so

$$\boxed{y=-\frac32x+10}.$$

Check $Q$:

$$-\frac32(8)+10=-12+10=-2.$$

Checking the second point catches most arithmetic and sign errors.

Self-check 2

Find the equation of the line through $(-4,-1)$ and $(2,8)$.

Answer

$$m=\frac{8-(-1)}{2-(-4)}=\frac96=\frac32.$$

Using $(2,8)$,

$$y-8=\frac32(x-2),$$

so

$$\boxed{y=\frac32x+5}.$$

Horizontal and vertical lines

A horizontal line has equation

$$\boxed{y=k},$$

because every point has the same $y$ coordinate. Its gradient is $0$.

A vertical line has equation

$$\boxed{x=k},$$

because every point has the same $x$ coordinate. Its gradient is undefined: calculating change in $y$ divided by change in $x$ would require division by zero.

A vertical line cannot be written as $y=mx+c$. It is not a function of $x$, because one $x$ value corresponds to many $y$ values.

Worked example 8: special cases

The line through $(4,-3)$ and $(4,9)$ is

$$\boxed{x=4}.$$

The line through $(-5,2)$ and $(7,2)$ is

$$\boxed{y=2}.$$

Do not try to force either result through the general two-point calculation without first noticing the equal coordinates.

Parallel and perpendicular lines

Parallel non-vertical lines have equal gradients:

$$\boxed{m_1=m_2}.$$

Distinct parallel lines have different intercepts. If both gradient and intercept are equal, the equations describe the same line.

Perpendicular lines meet at a right angle. For finite, non-zero gradients,

$$\boxed{m_1m_2=-1}.$$

Therefore the perpendicular gradient to $m$ is the negative reciprocal

$$-\frac1m.$$

One gradient changes sign and the fraction turns upside down. Horizontal and vertical lines form the exceptional perpendicular pair.

Worked example 9: a parallel line

Find the equation of the line parallel to $2x+5y=7$ through $(5,-1)$.

Rearrange the given line:

$$5y=-2x+7 \quad\Longrightarrow\quad y=-\frac25x+\frac75.$$

The required gradient is therefore $-\dfrac25$. Through $(5,-1)$:

$$y+1=-\frac25(x-5).$$

Hence

$$y+1=-\frac25x+2 \quad\Longrightarrow\quad \boxed{y=-\frac25x+1}.$$

Worked example 10: a perpendicular line

Find the equation of the perpendicular bisector of the segment joining $A(-2,3)$ and $B(6,7)$.

The midpoint is

$$M=\left(\frac{-2+6}{2},\frac{3+7}{2}\right)=(2,5).$$

The gradient of $AB$ is

$$m_{AB}=\frac{7-3}{6-(-2)}=\frac48=\frac12.$$

The perpendicular gradient is $-2$, since

$$\frac12(-2)=-1.$$

Using the midpoint:

$$y-5=-2(x-2),$$

so

$$\boxed{y=-2x+9}.$$

This line is perpendicular to $AB$ and passes through its midpoint, so both required properties have been established.

Common misconception: only change the sign

The perpendicular gradient to $3$ is $-\dfrac13$, not $-3$. The perpendicular gradient to $-\dfrac47$ is $\dfrac74$, not $\dfrac47$.

Intersections and simultaneous equations

At an intersection, the same point satisfies both line equations. If both equations give $y$, set their right sides equal.

Worked example 11: find an intersection

Find where

$$y=3x-4 \qquad\text{and}\qquad y=-x+8$$

intersect.

At the intersection,

$$3x-4=-x+8.$$

Therefore

$$4x=12 \quad\Longrightarrow\quad x=3.$$

Substitute into either equation:

$$y=3(3)-4=5.$$

The intersection is

$$\boxed{(3,5)}.$$

Substitution in the other equation gives $-3+8=5$, confirming the result. This is the graphical meaning of solving two simultaneous equations.

Straight lines as models

In a model

$$y=mx+c,$$

the gradient has units

$$\frac{\text{units of }y}{\text{units of }x},$$

and the intercept is the modelled value of $y$ when $x=0$. Interpret both in context.

Worked example 12: interpret a model

A taxi fare $C$ pounds is modelled by

$$C=2.4d+3.5,$$

where $d$ is the distance in kilometres.

• The gradient $2.4$ means the fare increases by £$2.40$ for each additional kilometre.
• The intercept $3.5$ represents the initial charge of £$3.50$.
• For a $7$ km journey,

$$C=2.4(7)+3.5=20.3,$$

so the model predicts a fare of £$20.30$.

The algebraic line extends indefinitely, but the practical model may only be sensible for distances in a stated domain, usually $d\geq0$.

A reliable method

When solving a straight-line problem:

1. Identify what is known: points, gradient, intercept, parallelism or perpendicularity.
2. Calculate the gradient carefully, preserving subtraction order.
3. Use $y-y_1=m(x-x_1)$ when a point and gradient are known.
4. Rearrange only if the requested form requires it.
5. Substitute known points to check the equation.
6. Treat horizontal and vertical lines separately.
7. Include units and interpretation in modelling questions.

Mixed self-check

1. State the gradient and $y$ intercept of $y=5-3x$.
2. Find both axis intercepts of $2x+3y=18$.
3. Find the equation of the line with gradient $\dfrac14$ through $(8,-3)$.
4. Find the equation through $(-1,6)$ and $(3,-2)$.
5. Find the equation of the line parallel to $3y=6x+7$ through $(4,1)$.
6. Find the equation of the line perpendicular to $y=-\dfrac35x+4$ through $(-3,2)$.
7. Find the intersection of $y=2x+1$ and $y=10-x$.
8. The line through $(a,4)$ and $(7,16)$ has gradient $3$. Find $a$.
9. Find the perpendicular bisector of the segment from $(1,-2)$ to $(5,6)$.
10. A tank contains $V=120-8t$ litres after $t$ minutes. Interpret the gradient and intercept, then find when the model predicts that the tank is empty.

Answers

1. $m=-3$ and $c=5$.
2. Setting $y=0$ gives $(9,0)$; setting $x=0$ gives $(0,6)$.
3. $y+3=\dfrac14(x-8)$, so $\boxed{y=\dfrac14x-5}$.
4. $m=\dfrac{-2-6}{3-(-1)}=-2$, so $\boxed{y=-2x+4}$.
5. The given gradient is $2$. Thus $y-1=2(x-4)$, so $\boxed{y=2x-7}$.
6. The perpendicular gradient is $\dfrac53$. Thus $y-2=\dfrac53(x+3)$, so $\boxed{y=\dfrac53x+7}$.
7. $2x+1=10-x$ gives $x=3$, then $y=7$. The intersection is $(3,7)$.
8. $3=\dfrac{16-4}{7-a}=\dfrac{12}{7-a}$, so $7-a=4$ and $\boxed{a=3}$.
9. The midpoint is $(3,2)$ and the segment gradient is $2$, so the perpendicular gradient is $-\dfrac12$. Hence $y-2=-\dfrac12(x-3)$, or $\boxed{y=-\dfrac12x+\dfrac72}$.
10. The initial volume is $120$ litres and the volume decreases by $8$ litres per minute. Setting $V=0$ gives $120-8t=0$, so the tank is empty after $15$ minutes.

What to learn next

Continue to graphs of functions to compare straight lines with other important graph families, and graph transformations to understand how equations move and reshape graphs.

For the full A-level treatment, including exact coordinate arguments and more demanding line problems, study straight lines in coordinate geometry. The same gradient ideas underpin differentiation and gradients.