1MA1 · A9–A10AlgebraFoundation & HigherGrades 4–7

Straight Line Graphs

Q: What is the gradient of a vertical line?

It is undefined, because the change in x is zero and you cannot divide by zero.

Q: How do I plot y = mx + c quickly?

Plot the intercept (0, c), then move 1 across and m up (or down) to get a second point. Use a third point as a check.

Every straight line on a coordinate grid can be described by an equation of the form $y = mx + c$ . The number $m$ says how steep the line is, and $c$ says where it crosses the $y$ -axis. Once you can move fluently between a line, its equation and the points on it, a whole family of questions — plotting, finding equations, parallel and perpendicular lines — becomes routine.

55 min Graphs

The big picture

The gradient measures how much $y$ changes for each step of 1 in $x$ . That makes it a rate: on a distance–time graph it is a speed, on a cost graph it is a price per item. Reading $m$ and $c$ in context is a common Edexcel question in its own right.

Parallel lines share a gradient; perpendicular lines have gradients whose product is $-1$ . That single fact about perpendicular gradients is a Higher-tier favourite and connects directly to tangents to circles later in this chapter.

What you'll learn

Plot straight line graphs from their equations
Find the gradient and $y$ -intercept from $y = mx + c$ and from a graph
Find the equation of a line through two points, or through a point with a given gradient
Identify and find parallel and perpendicular lines (perpendicular: Higher)

Gradient and intercept

In $y = mx + c$ , $m$ is the $\textbf{gradient}$ and $c$ is the $\textbf{y-intercept}$ , where the line crosses the $y$ -axis at $(0, c)$ .

A positive gradient slopes upwards from left to right; a negative gradient slopes downwards. A gradient of 0 gives a horizontal line $y = c$ ; vertical lines have the form $x = k$ .

If the equation is not in the form $y = mx + c$ , rearrange it first: $2y - 6x = 8$ becomes $y = 3x + 4$ .

y = mx + c

Gradient m, y-intercept c.

Your turnSame method, new numbers — try it before moving on.

What is the gradient of the line $3y = 6 - 2x$ ?

Your answer

Show worked solution

Solution

1Rearrange: $y = 2 - \frac{2}{3}x$ .
2The gradient is $-\frac{2}{3}$ .

Answer: $-\frac{2}{3}$

Tip — The gradient is the coefficient of $x$ only after $y$ is the subject. In $2y = 6x + 8$ the gradient is 3, not 6.

Gradient between two points

The gradient between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is the change in $y$ divided by the change in $x$ .

Subtract in the same order on the top and bottom. A drawn right-angled triangle on a graph gives the same value — rise over run.

m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}

Gradient between two points.

Example 1Find the equation of the line through

(-2, 5)

and

(4, -7)

m = \dfrac{-7 - 5}{4 - (-2)} = \dfrac{-12}{6} = -2

2Substitute

(-2, 5)

into

y = -2x + c

5 = 4 + c

, so

c = 1

Answer

y = -2x + 1

Your turnSame method, new numbers — try it before moving on.

Find the equation of the line through $(1, 3)$ and $(4, 12)$ .

Your answer

Type it like y = 2x + 1.

Show worked solution

Solution

1Gradient $= \frac{12 - 3}{4 - 1} = 3$ .
2 $3 = 3(1) + c$ , so $c = 0$ : $y = 3x$ .

Answer: $y = 3x$

Parallel and perpendicular lines

$\textbf{Parallel}$ lines have equal gradients: $y = 3x - 2$ and $y = 3x + 7$ are parallel.

$\textbf{Perpendicular}$ lines have gradients that multiply to $-1$ . If one gradient is $m$ , the other is $-\dfrac{1}{m}$ — the negative reciprocal.

To find a line perpendicular to a given one through a point, take the negative reciprocal gradient and substitute the point to find $c$ .

m_{1} \times m_{2} = -1

Perpendicular gradients.

Example 1Find the equation of the line perpendicular to

y = 3x - 4

that passes through

(6, 1)

1Perpendicular gradient:

-\tfrac13

2Substitute

(6, 1)

1 = -2 + c

, so

c = 3

Answer

y = -\tfrac13 x + 3

Your turnSame method, new numbers — try it before moving on.

Find the equation of the line perpendicular to $y = -2x + 5$ that passes through $(4, 1)$ .

Your answer

Type it like y = 1/2x - 1.

Show worked solution

Solution

1Perpendicular gradient: $\frac{1}{2}$ .
2 $1 = \frac{1}{2}(4) + c$ , so $c = -1$ .

Answer: $y = \frac{1}{2}x - 1$

Why this works

Rotating a line through $90^{\circ}$ swaps the rise and the run and changes one sign. A gradient of $\tfrac{3}{1}$ becomes $\tfrac{-1}{3}$ — which is exactly the negative reciprocal.

Lines in context

When a straight line models a real situation, $c$ is the starting value and $m$ is the rate of change.

A taxi fare of $F = 2.40d + 3.50$ pounds has a fixed charge of £3.50 and costs £2.40 per mile.

Example 1A plumber charges £45 for a 1-hour job and £105 for a 4-hour job. Assuming a linear charge, find the call-out fee and the hourly rate.

1Gradient:

\dfrac{105 - 45}{4 - 1} = 20

2Using

(1, 45)

45 = 20 + c

, so

c = 25

AnswerCall-out fee £25, plus £20 per hour.

Tip — In context questions, always interpret $m$ and $c$ with units: "£20 per hour", not just "20".

Think like an examiner

Show the gradient calculation with the coordinates substituted.
Show the substitution used to find $c$ .
Give equations in the form requested, such as $ax + by + c = 0$ with integers.

Straight lines

y = mx + c

Equation of a line.

m = \tfrac{y_{2} - y_{1}}{x_{2} - x_{1}}

Gradient.

m_{1}m_{2} = -1

Perpendicular lines.

Watch out for these

Computing

\dfrac{x_{2} - x_{1}}{y_{2} - y_{1}}

Change in

y

over change in

x

Taking the gradient of

3y = 6x + 2

as 6.

Make

y

the subject: gradient 2.

Using

-m

as the perpendicular gradient.

Use the negative reciprocal,

-\tfrac{1}{m}

Writing

y = 4

as a vertical line.

y = 4

is horizontal; vertical lines are

x = k

Stretch yourself

Points $A(1, 2)$ and $B(7, 6)$ are the ends of a line segment. Find the equation of the perpendicular bisector of $AB$ — the line through its midpoint at right angles to it.

Hint — Find the midpoint and the gradient of $AB$ , then use the negative reciprocal gradient through the midpoint.

Reveal solution

SolutionMidpoint

M = \left(\dfrac{1 + 7}{2}, \dfrac{2 + 6}{2}\right) = (4, 4)

. Gradient of

AB = \dfrac{6 - 2}{7 - 1} = \dfrac23

, so the perpendicular gradient is

-\dfrac32

. Through

(4, 4)

4 = -6 + c

, so

c = 10

. The perpendicular bisector is

y = -\tfrac32 x + 10

, or

3x + 2y = 20

. Check: it passes through

(4, 4)

since

12 + 8 = 20

✓.

\blacksquare

Questions students ask

What is the gradient of a vertical line?

It is undefined, because the change in

x

is zero and you cannot divide by zero.

How do I plot

y = mx + c

quickly?

Plot the intercept

(0, c)

, then move 1 across and

m

up (or down) to get a second point. Use a third point as a check.

Do perpendicular lines always have one positive and one negative gradient?

Yes, unless one is horizontal and the other vertical.

Key takeaways

$y = mx + c$ : gradient $m$ , $y$ -intercept $(0, c)$ .
Gradient $= \tfrac{\text{change in } y}{\text{change in } x}$ .
Find $c$ by substituting a known point.
Parallel lines: equal gradients. Perpendicular lines: $m_{1}m_{2} = -1$ .
In context, $m$ is a rate and $c$ a starting value.

Exam-style questions

Practise it the way it is examined

3 original questions · 7 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 42 marks

Find the gradient and $y$ -intercept of the line $2y - 4x = 10$ .[2]
Show mark scheme
- M1Rearranges to $y = 2x + 5$
- A1Gradient 2, $y$ -intercept 5

Question 2Grade 62 marks

Line $L$ passes through $(2, 7)$ and is parallel to $y = 3x - 1$ . Find the equation of $L$ .[2]
Show mark scheme
- M1Uses gradient 3 with $(2, 7)$
- A1 $y = 3x + 1$
Answer: $y = 3x + 1$

Question 3Grade 73 marks

$A$ is $(-1, 4)$ and $B$ is $(5, 2)$ . Find the equation of the perpendicular bisector of $AB$ .[3]
Show mark scheme
- M1Midpoint $(2, 3)$
- M1Gradient of $AB = -\frac{1}{3}$ , so perpendicular gradient $3$
- A1 $y = 3x - 3$
Answer: $y = 3x - 3$

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