Chapter 12 · Higher tier

GCSE Maths Straight-Line Graphs and Coordinates (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

Plot and read coordinates in all four quadrants, and find missing corners of shapes
Find the midpoint of a line segment and its length using Pythagoras’ theorem
Draw straight-line graphs from tables of values, including horizontal and vertical lines
Find and interpret the gradient and y-intercept of a line, and use y = mx + c
Find the equation of a line from a graph, two points, or a point and a gradient
Use gradients to identify parallel and (Higher) perpendicular lines
Draw and interpret real-life graphs, and read the gradient as a rate of change
Find where two lines cross

You should already know

Negative numbers and the four operations (Chapter 1)
Substituting into expressions and formulae (Chapter 8)
Solving linear equations (Chapter 9)
Simplifying fractions (Chapter 3)

Where this leads

Quadratic and other graphs (Chapter 13)
Simultaneous equations (Chapter 14)
Direct proportion and compound measures (Chapter 16)
Pythagoras’ theorem in full (Chapter 20)
Transformations on a grid (Chapter 21)
Functions and gradients of curves (Chapter 25)

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: A8, A9, A10, A14, G11, R14.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

12.1 Coordinates in four quadrants: Worksheet (PDF) · Answers (PDF) · lesson page
12.2 Midpoints and lengths of line segments: Worksheet (PDF) · Answers (PDF) · lesson page
12.3 Horizontal and vertical lines, and drawing graphs from tables: Worksheet (PDF) · Answers (PDF) · lesson page
12.4 Gradient: Worksheet (PDF) · Answers (PDF) · lesson page
12.5 The equation y = mx + c: Worksheet (PDF) · Answers (PDF) · lesson page
12.6 Finding the equation of a straight line: Worksheet (PDF) · Answers (PDF) · lesson page
12.7 Parallel and perpendicular lines: Worksheet (PDF) · Answers (PDF) · lesson page
12.8 Real-life graphs and rates of change: Worksheet (PDF) · Answers (PDF) · lesson page
12.9 Where lines cross: Worksheet (PDF) · Answers (PDF) · lesson page
Check-up and exam-style questions: Worksheet (PDF) · Answers (PDF) · lesson page

Exercise P Do you remember?

Non-calculator

FWork out 5 − (−3).

Done

FWork out the value of 2x + 1 when x = −3.

Done

FSolve 3x + 2 = 14.

Done

FSimplify the fraction −6/4 (write it as a fraction in its simplest form).

Done

FA right-angled triangle has two short sides of 6 cm and 8 cm. Work out the length of the longest side.

Done

12.1

Coordinates in four quadrants

Edexcel 1MA1: A8, G11
Grade 1-3

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Coordinates in four quadrantsComing soon

A coordinate grid lets us give the exact position of any point using two numbers. It is made from two number lines that cross at right angles. The horizontal one is the x-axis and the vertical one is the y-axis. They cross at the origin, which is the point (0, 0).

A point is described by a pair of numbers in brackets, called its coordinates: (x, y). The first number says how far to go across from the origin, and the second says how far to go up or down. A good way to remember the order is “along the corridor, then up the stairs”: you cannot go up before you have gone along. In the alphabet x comes before y, which gives the same order.

Numbers on the axes carry on past zero into the negatives. Going right or up is positive. Going left or down is negative. So the point (−4, 3) is 4 squares to the left of the origin and 3 squares up.

The axes split the grid into four quadrants, numbered 1 to 4 going anticlockwise from the top right.

The four regions are called quadrants. They are numbered anticlockwise, starting at the top right. The signs of the coordinates tell you which quadrant a point is in: in Quadrant 1 both coordinates are positive; in Quadrant 3 both are negative. A point on an axis has one coordinate equal to 0, so it is not inside any quadrant. Points on the y-axis have x = 0, for example (0, 4); points on the x-axis have y = 0, for example (−3, 0).

Coordinates and shapes

Corners of shapes are called vertices (one is a vertex). On a grid, points on a horizontal line all have the same y-coordinate, and points on a vertical line all have the same x-coordinate. This lets you find missing corners of rectangles, squares and parallelograms, and work out lengths by subtracting. The distance from −3 to 5 along an axis is 5 − (−3) = 8: always take the smaller number from the larger.

Key ideas

The coordinates (x, y) are written across first, then up or down.
The origin is (0, 0). Right and up are positive; left and down are negative.
Quadrants are numbered 1, 2, 3, 4 anticlockwise from the top right: (+, +), (−, +), (−, −), (+, −).
A point on an axis has one coordinate equal to 0.
In a rectangle, opposite sides are equal and parallel; in a parallelogram too. Use this to find a missing vertex.

Worked example 12.1.1

Write down the coordinates of the points P, Q, R and S.

Key words. A coordinate pair such as (3, 2) gives the position of a point. The first number is the x-coordinate (how far across, along the horizontal axis); the second is the y-coordinate (how far up or down). The origin is the point (0, 0) where the axes cross.

Method. For each point go across from the origin first, then up or down. Write the coordinates as (x, y).

Point P: how far across? how far up or down? Across means along the x-axis. Right is positive.
P is 3 right and 2 up: P = (3, 2)
Point Q: left of the origin. Left is negative, so the x-coordinate is negative.
Q is 4 left and 1 up: Q = (−4, 1)
Point R: left and down. Both coordinates are negative.
R is 2 left and 3 down: R = (−2, −3)
Point S: it is on the y-axis. It is directly above the origin, so there is no movement across: x = 0.
S is 0 across and 4 up: S = (0, 4)

Check. Start at (0, 0) and follow each pair of numbers. You arrive at the right point each time. ✓

Watch out. Across first, then up or down: (3, 2) is not the same as (2, 3).

Worked example 12.1.2

ABCD is a rectangle. A is (−4, −2), B is (3, −2) and C is (3, 1). Find the coordinates of D.

Method. In a rectangle, opposite sides are parallel and equal, so the move from B to A is the same as the move from C to D.

Plot the three points A, B and C. A sketch helps you see where the fourth corner has to go. The corners are labelled in order round the shape.
A = (−4, −2), B = (3, −2), C = (3, 1)
Work out the move from B to A. In a rectangle, opposite sides are parallel and equal in length, and every angle is 90°. So side CD is the same move as side BA.
across: −4 − 3 = −7 (7 left)
up/down: −2 − (−2) = 0 (0)
Start at C and make the same move to reach D. C is at (3, 1). Add the same across and up/down numbers.
x: 3 + (−7) = −4
y: 1 + 0 = 1
Write the coordinates and draw the shape.
D = (−4, 1)

Check. The diagonals of a rectangle cross at their midpoints. Midpoint of AC = (−0.5, −0.5) and midpoint of BD = (−0.5, −0.5). They are the same. ✓

Watch out. Keep the corners in order round the shape: A, B, C, D. If you join them in the wrong order you get a bow-tie, not a parallelogram.

Worked example 12.1.3

ABCD is a parallelogram. A is (−3, −1), B is (2, −3) and C is (6, −3). Find the coordinates of D.

Method. In a parallelogram, opposite sides are parallel and equal, so the move from B to A is the same as the move from C to D.

Plot the three points A, B and C. A sketch helps you see where the fourth corner has to go. The corners are labelled in order round the shape.
A = (−3, −1), B = (2, −3), C = (6, −3)
Work out the move from B to A. In a parallelogram, opposite sides are parallel and equal in length. So side CD is the same move as side BA.
across: −3 − 2 = −5 (5 left)
up/down: −1 − (−3) = 2 (2 up)
Start at C and make the same move to reach D. C is at (6, −3). Add the same across and up/down numbers.
x: 6 + (−5) = 1
y: −3 + 2 = −1
Write the coordinates and draw the shape.
D = (1, −1)

Check. The diagonals of a parallelogram cross at their midpoints. Midpoint of AC = (1.5, −2) and midpoint of BD = (1.5, −2). They are the same. ✓

Watch out. Keep the corners in order round the shape: A, B, C, D. If you join them in the wrong order you get a bow-tie, not a parallelogram.

Worked example 12.1.4

A triangle has vertices A(−4, 1), B(2, 1) and C(2, −3). Work out its area.

Method. Find the base and height from the coordinates, then area of a triangle = ½ × base × height.

Sketch the triangle. Plot the three points and join them.
Find the base and the height. Two sides meet at a right angle at B, one horizontal and one vertical. These are the base and the height. Subtract coordinates to find each length.
base = 6, height = 4
Use area = ½ × base × height. The triangle is half of a rectangle that is base by height.
6 × 4 ÷ 2 = 24 ÷ 2 = 12

Check. The rectangle round the triangle has area 6 × 4 = 24; half of it is 12. ✓

Watch out. Use the perpendicular height, not a slanted side.

Watch out

Order matters. (3, 2) and (2, 3) are different points. Also take care with minus signs: (−2, 5) is left of the y-axis, so it is in Quadrant 2, not Quadrant 4. When you count the distance between points, use bigger − smaller, and remember 5 − (−3) = 8.

12.2

Midpoints and lengths of line segments

Edexcel 1MA1: G11, A8
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Midpoints and lengths of line segmentsComing soon

A line segment is the part of a straight line between two points. Two questions come up all the time: where is its middle? and how long is it?

The midpoint

The midpoint is the point exactly halfway between the two ends. To find it, work on the x-coordinates and the y-coordinates separately. The number halfway between two numbers is their mean: add them and divide by 2. For example, halfway between 2 and 8 is (2 + 8) ÷ 2 = 5, because 5 is 3 above 2 and 3 below 8. So for ends (x₁, y₁) and (x₂, y₂):

midpoint = ( (x₁ + x₂) ÷ 2 , (y₁ + y₂) ÷ 2 )

The midpoint M of the line from A(−4, 1) to B(6, 5) is (1, 3): halfway across and halfway up.

The length of a line segment

To find the length of a sloping line, make it the longest side of a right-angled triangle. Draw a horizontal line across and a vertical line up (or down) between the two points. The lengths of these two short sides come from subtracting the coordinates. Then use Pythagoras’ theorem: in a right-angled triangle, the square of the longest side (the hypotenuse) equals the sum of the squares of the other two sides, c² = a² + b². You will meet this theorem in full in Chapter 20; here we use it for a single job.

AB is the hypotenuse of a right-angled triangle with sides 6 across and 8 up: AB² = 6² + 8² = 100, so AB = 10.

Some triangles have whole-number sides, such as 3, 4, 5 and 5, 12, 13 and 6, 8, 10, and the length is an exact whole number. Otherwise the answer is a decimal on your calculator (round it as asked), or on the Higher tier you may be asked to leave it as a surd, such as √13 or 2√13.

Working backwards (Higher)

If you know the midpoint M and one end A, the other end B is the same step again beyond M. If M is 3 across and 2 up from A, then B is another 3 across and 2 up from M. Lengths and midpoints are also used to prove things about shapes: for example, a triangle with two equal sides is isosceles, and the centre of a circle is the midpoint of any diameter.

Key ideas

Midpoint of (x₁, y₁) and (x₂, y₂) = ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2): add then halve.
Length: find the horizontal distance and the vertical distance (subtract), then Pythagoras: length² = (across)² + (up)².
Square first, add, then square root last: √(a² + b²) is not a + b.
Higher: leave exact answers as surds (for example 2√13); to find an end point, repeat the step from the known end to the midpoint.

Worked example 12.2.1

Find the midpoint of A(2, 3) and B(8, 7).

Key words. The midpoint of a line segment is the point exactly halfway along it. The mean of two numbers is their sum divided by 2.

Method. The midpoint is exactly halfway between two points. Find the mean of the x-coordinates and the mean of the y-coordinates.

Write down the two points and pick out their coordinates. The midpoint uses the x-coordinates together and the y-coordinates together.
A = (2, 3): x₁ = 2, y₁ = 3
B = (8, 7): x₂ = 8, y₂ = 7
Find the x-coordinate of the midpoint. The number halfway between two numbers is their mean: add them and divide by 2.
(2 + 8) ÷ 2 = 10 ÷ 2 = 5
Find the y-coordinate of the midpoint. Do exactly the same with the y-coordinates.
(3 + 7) ÷ 2 = 10 ÷ 2 = 5
Write the answer as a coordinate pair. Across first, then up or down.
Midpoint = (5, 5)
Sketch to see that it makes sense. The midpoint M must be on the line, exactly halfway between the two ends.

Check. From A to the midpoint the x-coordinate changes by 3 and from the midpoint to B it changes by 3; the y-coordinate changes by 2 and 2. The two steps are equal, so M is halfway. ✓

Watch out. Add the coordinates and halve. Subtracting gives the length of the gap, not the halfway point. Work with x and y separately.

Worked example 12.2.2

Find the midpoint of P(−4, 1) and Q(3, 6).

Key words. The midpoint of a line segment is the point exactly halfway along it. The mean of two numbers is their sum divided by 2.

Method. The midpoint is exactly halfway between two points. Find the mean of the x-coordinates and the mean of the y-coordinates.

Write down the two points and pick out their coordinates. The midpoint uses the x-coordinates together and the y-coordinates together.
A = (−4, 1): x₁ = −4, y₁ = 1
B = (3, 6): x₂ = 3, y₂ = 6
Find the x-coordinate of the midpoint. The number halfway between two numbers is their mean: add them and divide by 2.
(−4 + 3) ÷ 2 = −1 ÷ 2 = −0.5
Find the y-coordinate of the midpoint. Do exactly the same with the y-coordinates.
(1 + 6) ÷ 2 = 7 ÷ 2 = 3.5
Write the answer as a coordinate pair. Across first, then up or down.
Midpoint = (−0.5, 3.5)
Sketch to see that it makes sense. The midpoint M must be on the line, exactly halfway between the two ends.

Check. From A to the midpoint the x-coordinate changes by 3.5 and from the midpoint to B it changes by 3.5; the y-coordinate changes by 2.5 and 2.5. The two steps are equal, so M is halfway. ✓

Watch out. Add the coordinates and halve. Subtracting gives the length of the gap, not the halfway point. Work with x and y separately.

Worked example 12.2.3

Work out the length of the line segment from A(1, 2) to B(7, 10).

Key words. A line segment is the part of a line between two points. Pythagoras’ theorem says that in a right-angled triangle the square of the hypotenuse equals the sum of the squares of the other two sides (see Chapter 20).

Method. Draw a right-angled triangle using the horizontal and vertical distances between the points, then use Pythagoras’ theorem to find the length of the line (the hypotenuse).

Find the horizontal distance between the points (the difference in x). Subtract the x-coordinates and ignore the sign: a distance cannot be negative.
7 − 1 = 6
Find the vertical distance (the difference in y). Do the same with the y-coordinates.
10 − 2 = 8
Sketch a right-angled triangle on the line. The line AB is the hypotenuse (the longest side, opposite the right angle). The two shorter sides are the horizontal and vertical distances.
Use Pythagoras’ theorem: c² = a² + b². Square the two shorter sides and add them to get the hypotenuse squared.
AB² = 6² + 8² = 36 + 64 = 100
Take the square root. The square root undoes the squaring, giving the length itself.
AB = √100 = 10

Check. 10² = 100 = 100. ✓

Watch out. Add the squares before square-rooting. √(9 + 16) is 5, not 3 + 4.

Worked example 12.2.4

Work out the length of the line segment from P(−3, −1) to Q(2, 4). Give your answer correct to 1 decimal place.

Method. Draw a right-angled triangle using the horizontal and vertical distances between the points, then use Pythagoras’ theorem to find the length of the line (the hypotenuse).

Find the horizontal distance between the points (the difference in x). Subtract the x-coordinates and ignore the sign: a distance cannot be negative.
2 − (−3) = 5
Find the vertical distance (the difference in y). Do the same with the y-coordinates.
4 − (−1) = 5
Sketch a right-angled triangle on the line. The line AB is the hypotenuse (the longest side, opposite the right angle). The two shorter sides are the horizontal and vertical distances.
Use Pythagoras’ theorem: c² = a² + b². Square the two shorter sides and add them to get the hypotenuse squared.
AB² = 5² + 5² = 25 + 25 = 50
Take the square root and round. The square root undoes the squaring. Use a calculator, then round as the question asks.
AB = √50 = 7.0711...
= 7.1 to 1 decimal place

Check. The line should be longer than the longer leg (5) and shorter than the two legs added (10). 7.1 is in between. ✓

Watch out. Add the squares before square-rooting. √(9 + 16) is 5, not 3 + 4.

Worked example 12.2.5

M(3, 2) is the midpoint of AB. A is (1, −4). Find the coordinates of B.

▶ Watch this solution as a video

Method. The midpoint is halfway, so the other end is the same distance beyond the midpoint. Alternatively use the midpoint rule backwards: 2 × midpoint − known end.

Write what we know. A is one end and M is the midpoint. The other end is B = (x, y), which we want to find.
A = (1, −4), M = (3, 2)
Work out how far the midpoint is from the known end. The midpoint is halfway, so the other end is the same step again beyond M.
Across: 3 − 1 = 2
Up/down: 2 − (−4) = 6
Add the same step again to the midpoint. Going from A to M is half of the journey to B, so repeat it once more starting at M.
x: 3 + 2 = 5
y: 2 + 6 = 8
Write the coordinates.
B = (5, 8)

Check. Midpoint of (1, −4) and (5, 8): (1 + 5) ÷ 2 = 3 and (−4 + 8) ÷ 2 = 2. This is M. ✓

Watch out. Do not just add or subtract the midpoint and the end. The step from the end to M must be repeated once more.

Worked example 12.2.6

Find the exact length of the line from A(−2, 1) to B(4, 5). Give your answer as a simplified surd.

▶ Watch this solution as a video

Method. Draw a right-angled triangle using the horizontal and vertical distances between the points, then use Pythagoras’ theorem to find the length of the line (the hypotenuse).

Find the horizontal distance between the points (the difference in x). Subtract the x-coordinates and ignore the sign: a distance cannot be negative.
4 − (−2) = 6
Find the vertical distance (the difference in y). Do the same with the y-coordinates.
5 − 1 = 4
Sketch a right-angled triangle on the line. The line AB is the hypotenuse (the longest side, opposite the right angle). The two shorter sides are the horizontal and vertical distances.
Use Pythagoras’ theorem: c² = a² + b². Square the two shorter sides and add them to get the hypotenuse squared.
AB² = 6² + 4² = 36 + 16 = 52
Take the square root. This gives the exact length as a surd. A surd is a square root that is not a whole number.
AB = √52
Simplify the surd. Look for a square number that divides 52: 52 = 4 × 13, and √4 = 2.
√52 = √4 × √13 = 2√13

Check. On a calculator √52 = 7.211, which matches 2√13 = 7.211. A sketch agrees: the line is longer than 6 but shorter than 10. ✓

Watch out. Add the squares before square-rooting. √(9 + 16) is 5, not 3 + 4.

Watch out

Add for the midpoint, subtract for the distance. A common error is to subtract the coordinates when asked for the midpoint. Another is to write √(9 + 16) as 3 + 4: you must add the squares before taking the square root, giving √25 = 5.

12.3

Horizontal and vertical lines, and drawing graphs from tables

Edexcel 1MA1: A9
Grade 2-4

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

An equation of a line is a rule that every point on the line obeys. Plot the points (3, −2), (3, 0) and (3, 4). They all have x = 3, and they line up straight up and down. The rule “x is always 3” is written x = 3, and it is the equation of a vertical line that passes through 3 on the x-axis. In the same way y = −2 is a horizontal line passing through −2 on the y-axis, because every point on it has y = −2.

Two special lines are the axes themselves. The y-axis is the line x = 0 and the x-axis is the line y = 0.

Vertical lines are x = a number. Horizontal lines are y = a number.

Drawing a line from its equation

Most lines are neither flat nor upright. A line such as y = 2x + 1 is a rule that tells you what y is for any x: double the x, then add 1. To draw it:

Make a table of values. Choose some x values (three or four, including negatives and 0) and work out y for each.
Plot the points (x, y) on a grid.
Join them with a ruler to make a straight line, and extend it across the grid. Add the equation as a label.

Why does this give a straight line? Because x appears only on its own (not as x² or 1/x), so each time x goes up by 1, y always goes up by the same amount. Equations like this are called linear. You only need two points to draw a straight line, but plot at least three: if they do not line up, you have made a mistake in your working.

When you substitute a negative number, put it in brackets: if x = −2 in y = 2x + 1 write 2 × (−2) + 1 = −4 + 1 = −3. A point is on the line if its coordinates make the equation true. For example (3, 7) is on y = 2x + 1 because 2 × 3 + 1 = 7, but (3, 8) is not.

The graph of y = 2x + 1 drawn from the table of values x = −2, 0, 1, 3.

Key ideas

x = a is a vertical line; y = b is a horizontal line. The axes are y = 0 (the x-axis) and x = 0 (the y-axis).
To draw a line: table of values → plot the points → ruler → label.
Substitute negatives with brackets: 2 × (−3) = −6.
A point is on a line only if its coordinates make the equation true.
Check that all your points line up: if one does not, recheck its y value.

Worked example 12.3.1

Write down the equation of the vertical line labelled A in the diagram.

Key words. A line is horizontal if it is flat, like the horizon, and vertical if it goes straight up and down. The equation of a line is a rule that every point on the line obeys.

Method. A vertical (up and down) line has one coordinate that never changes. The equation says what that coordinate is.

Pick three points on the line. Choose points where the line crosses the grid lines so they are easy to read.
(2, −2), (2, 1), (2, 3)
Look for the coordinate that does not change. Every point on a vertical line is directly above or below the others, so they share the same x-coordinate.
The x-coordinate is 2 every time. The y-coordinate changes.
Write the equation. The equation says what is true for every point on the line: x is always 2.
x = 2

Check. It crosses the x-axis at (2, 0). ✓

Watch out. Vertical lines have equations x = a number; horizontal lines have equations y = a number. Test one point: (3, 0) is on x = 3 but not on y = 3.

Worked example 12.3.2

Draw the graph of y = 2x − 1 for values of x from −2 to 3.

Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).

Method. Draw a table of values, plot the points, then join them with a straight line.

Make a table of values. Two points are enough to draw a straight line, but three or more let you spot a mistake: all the points must line up.
Work out y for each x. Substitute each x into the equation.
x = −2: y = 2 × (−2) − 1 = −4 − 1 = −5
x = −1: y = 2 × (−1) − 1 = −2 − 1 = −3
x = 0: y = 2 × 0 − 1 = 0 − 1 = −1
x = 1: y = 2 × 1 − 1 = 2 − 1 = 1
x = 2: y = 2 × 2 − 1 = 4 − 1 = 3
x = 3: y = 2 × 3 − 1 = 6 − 1 = 5
x −2 −1 0 1 2 3
y −5 −3 −1 1 3 5
Plot the points. Each pair is a coordinate: across first, then up or down.
(−2, −5) (−1, −3) (0, −1) (1, 1) (2, 3) (3, 5)
Join them with a ruler and extend the line. The graph of a linear equation is a straight line, so the points must line up. If one is out of line, check its working.

x	−2	−1	0	1	2	3
y	−5	−3	−1	1	3	5

Check. All the plotted points lie on one straight line. A point taken from the line, such as one with a different x, also satisfies the equation. ✓

Watch out. Use a ruler and extend the line across the whole grid. If the points do not line up, one y value is wrong.

Worked example 12.3.3

Draw the graph of y = −x + 3 for values of x from −2 to 4. (Use x = −2, 0, 2, 4.)

Method. Draw a table of values, plot the points, then join them with a straight line.

Make a table of values. Two points are enough to draw a straight line, but three or more let you spot a mistake: all the points must line up.
Work out y for each x. Substitute each x into the equation.
x = −2: y = −(−2) + 3 = 2 + 3 = 5
x = 0: y = −0 + 3 = 0 + 3 = 3
x = 2: y = −2 + 3 = −2 + 3 = 1
x = 4: y = −4 + 3 = −4 + 3 = −1
x −2 0 2 4
y 5 3 1 −1
Plot the points. Each pair is a coordinate: across first, then up or down.
(−2, 5) (0, 3) (2, 1) (4, −1)
Join them with a ruler and extend the line. The graph of a linear equation is a straight line, so the points must line up. If one is out of line, check its working.

x	−2	0	2	4
y	5	3	1	−1

Check. All the plotted points lie on one straight line. A point taken from the line, such as one with a different x, also satisfies the equation. ✓

Watch out. Use a ruler and extend the line across the whole grid. If the points do not line up, one y value is wrong.

Worked example 12.3.4

Is the point (4, 10) on the line y = 3x − 2?

Method. A point is on a line if its coordinates make the equation true. Substitute them in.

Substitute the x-coordinate into the equation. If the point is on the line, the equation gives its y-coordinate.
y = 3 × 4 − 2 = 10
Compare with the y-coordinate of the point. The point is (4, 10), so its y-coordinate is 10.
10 = 10
Write the conclusion. The point is on the line only if the two values are equal.
(4, 10) is on the line y = 3x − 2.

Check. Try the substitution the other way: put the y-coordinate in and see if you get the x-coordinate. ✓

Watch out. Put the x in the x place and the y in the y place.

Watch out

Vertical or horizontal? x = 3 is vertical (it goes through 3 on the x-axis) and y = 3 is horizontal. If you are unsure, plot (3, 0) and (0, 3) and see which one is on the line. Also: when x is negative, put it in brackets before you multiply, or you may lose the minus sign.

Exercise 12D Horizontal and vertical lines, tables of values and drawing lines

Non-calculator

FWrite down the equation of each of the lines A, B, C and D.

(a)line A

(b)line B

(c)line C

(d)line D

Done

FWrite down the equation of each line.

(a)The line is vertical and passes through (−2, 4).

(b)The line is horizontal and passes through (3, 5).

(c)The line is the x-axis.

(d)The line is the y-axis.

Done

FComplete the table of values for each equation.

(a)y = x + 1 for x = 0, 1, 2, 3, 4

Key words. A table of values lists some x values and the matching y values. To substitute means to replace a letter with a number.

Method. Put each value of x into the equation and work out y. Do the multiplication before the addition.

Write the rule. The equation tells us how to get y from x: multiply by the gradient, then add the constant.
y = x + 1
Substitute each x value in turn. Replace the letter x with the number. Put negative numbers in brackets so you do not lose the sign.
x = 0: y = 0 + 1 = 1
x = 1: y = 1 + 1 = 2
x = 2: y = 2 + 1 = 3
x = 3: y = 3 + 1 = 4
x = 4: y = 4 + 1 = 5
Write the answers in the table. Each pair (x, y) is a point on the line.
x 0 1 2 3 4
y 1 2 3 4 5

x	0	1	2	3	4
y	1	2	3	4	5

Check. The y-values change by the same amount each time (the gradient 1 times the step in x): 1, 1, 1, 1. ✓

Watch out. Careful with negatives: 2 × (−3) = −6, and −6 + 1 = −5. A negative times a positive is negative.

(b)y = 2x for x = −2, −1, 0, 1, 2

Key words. A table of values lists some x values and the matching y values. To substitute means to replace a letter with a number.

Method. Put each value of x into the equation and work out y. Do the multiplication before the addition.

Write the rule. The equation tells us how to get y from x: multiply by the gradient, then add the constant.
y = 2x
Substitute each x value in turn. Replace the letter x with the number. Put negative numbers in brackets so you do not lose the sign.
x = −2: y = 2 × (−2) = −4
x = −1: y = 2 × (−1) = −2
x = 0: y = 2 × 0 = 0
x = 1: y = 2 × 1 = 2
x = 2: y = 2 × 2 = 4
Write the answers in the table. Each pair (x, y) is a point on the line.
x −2 −1 0 1 2
y −4 −2 0 2 4

x	−2	−1	0	1	2
y	−4	−2	0	2	4

Check. The y-values change by the same amount each time (the gradient 2 times the step in x): 2, 2, 2, 2. ✓

Watch out. Careful with negatives: 2 × (−3) = −6, and −6 + 1 = −5. A negative times a positive is negative.

(c)y = 3x − 1 for x = −1, 0, 1, 2, 3

Key words. A table of values lists some x values and the matching y values. To substitute means to replace a letter with a number.

Method. Put each value of x into the equation and work out y. Do the multiplication before the addition.

Write the rule. The equation tells us how to get y from x: multiply by the gradient, then add the constant.
y = 3x − 1
Substitute each x value in turn. Replace the letter x with the number. Put negative numbers in brackets so you do not lose the sign.
x = −1: y = 3 × (−1) − 1 = −3 − 1 = −4
x = 0: y = 3 × 0 − 1 = 0 − 1 = −1
x = 1: y = 3 × 1 − 1 = 3 − 1 = 2
x = 2: y = 3 × 2 − 1 = 6 − 1 = 5
x = 3: y = 3 × 3 − 1 = 9 − 1 = 8
Write the answers in the table. Each pair (x, y) is a point on the line.
x −1 0 1 2 3
y −4 −1 2 5 8

x	−1	0	1	2	3
y	−4	−1	2	5	8

Check. The y-values change by the same amount each time (the gradient 3 times the step in x): 3, 3, 3, 3. ✓

Watch out. Careful with negatives: 2 × (−3) = −6, and −6 + 1 = −5. A negative times a positive is negative.

(d)y = −2x + 10 for x = 0, 1, 2, 3, 4

Key words. A table of values lists some x values and the matching y values. To substitute means to replace a letter with a number.

Method. Put each value of x into the equation and work out y. Do the multiplication before the addition.

Write the rule. The equation tells us how to get y from x: multiply by the gradient, then add the constant.
y = −2x + 10
Substitute each x value in turn. Replace the letter x with the number. Put negative numbers in brackets so you do not lose the sign.
x = 0: y = (−2) × 0 + 10 = 0 + 10 = 10
x = 1: y = (−2) × 1 + 10 = −2 + 10 = 8
x = 2: y = (−2) × 2 + 10 = −4 + 10 = 6
x = 3: y = (−2) × 3 + 10 = −6 + 10 = 4
x = 4: y = (−2) × 4 + 10 = −8 + 10 = 2
Write the answers in the table. Each pair (x, y) is a point on the line.
x 0 1 2 3 4
y 10 8 6 4 2

x	0	1	2	3	4
y	10	8	6	4	2

Check. The y-values change by the same amount each time (the gradient −2 times the step in x): −2, −2, −2, −2. ✓

Watch out. Careful with negatives: 2 × (−3) = −6, and −6 + 1 = −5. A negative times a positive is negative.

Done

F(a) Complete a table of values for y = 2x + 1 for x = −2, −1, 0, 1, 2, 3. (b) Draw the graph of y = 2x + 1.

Method. Draw a table of values, plot the points, then join them with a straight line.

Make a table of values. Two points are enough to draw a straight line, but three or more let you spot a mistake: all the points must line up.
Work out y for each x. Substitute each x into the equation.
x = −2: y = 2 × (−2) + 1 = −4 + 1 = −3
x = −1: y = 2 × (−1) + 1 = −2 + 1 = −1
x = 0: y = 2 × 0 + 1 = 0 + 1 = 1
x = 1: y = 2 × 1 + 1 = 2 + 1 = 3
x = 2: y = 2 × 2 + 1 = 4 + 1 = 5
x = 3: y = 2 × 3 + 1 = 6 + 1 = 7
x −2 −1 0 1 2 3
y −3 −1 1 3 5 7
Plot the points. Each pair is a coordinate: across first, then up or down.
(−2, −3) (−1, −1) (0, 1) (1, 3) (2, 5) (3, 7)
Join them with a ruler and extend the line. The graph of a linear equation is a straight line, so the points must line up. If one is out of line, check its working.

x	−2	−1	0	1	2	3
y	−3	−1	1	3	5	7

Check. All the plotted points lie on one straight line. A point taken from the line, such as one with a different x, also satisfies the equation. ✓

Watch out. Use a ruler and extend the line across the whole grid. If the points do not line up, one y value is wrong.

Done

FDraw the graph of y = x − 2 for values of x from −3 to 4.

Method. Draw a table of values, plot the points, then join them with a straight line.

Make a table of values. Two points are enough to draw a straight line, but three or more let you spot a mistake: all the points must line up.
Work out y for each x. Substitute each x into the equation.
x = −3: y = (−3) − 2 = −5
x = −2: y = (−2) − 2 = −4
x = −1: y = (−1) − 2 = −3
x = 0: y = 0 − 2 = −2
x = 1: y = 1 − 2 = −1
x = 2: y = 2 − 2 = 0
x = 3: y = 3 − 2 = 1
x = 4: y = 4 − 2 = 2
x −3 −2 −1 0 1 2 3 4
y −5 −4 −3 −2 −1 0 1 2
Plot the points. Each pair is a coordinate: across first, then up or down.
(−3, −5) (−2, −4) (−1, −3) (0, −2) (1, −1) (2, 0) (3, 1) (4, 2)
Join them with a ruler and extend the line. The graph of a linear equation is a straight line, so the points must line up. If one is out of line, check its working.

x	−3	−2	−1	0	1	2	3	4
y	−5	−4	−3	−2	−1	0	1	2

Check. All the plotted points lie on one straight line. A point taken from the line, such as one with a different x, also satisfies the equation. ✓

Watch out. Use a ruler and extend the line across the whole grid. If the points do not line up, one y value is wrong.

Done

FDraw the graph of y = 3 − x for values of x from −2 to 4.

Method. Draw a table of values, plot the points, then join them with a straight line.

Make a table of values. Two points are enough to draw a straight line, but three or more let you spot a mistake: all the points must line up.
Work out y for each x. Substitute each x into the equation.
x = −2: y = −(−2) + 3 = 2 + 3 = 5
x = −1: y = −(−1) + 3 = 1 + 3 = 4
x = 0: y = −0 + 3 = 0 + 3 = 3
x = 1: y = −1 + 3 = −1 + 3 = 2
x = 2: y = −2 + 3 = −2 + 3 = 1
x = 3: y = −3 + 3 = −3 + 3 = 0
x = 4: y = −4 + 3 = −4 + 3 = −1
x −2 −1 0 1 2 3 4
y 5 4 3 2 1 0 −1
Plot the points. Each pair is a coordinate: across first, then up or down.
(−2, 5) (−1, 4) (0, 3) (1, 2) (2, 1) (3, 0) (4, −1)
Join them with a ruler and extend the line. The graph of a linear equation is a straight line, so the points must line up. If one is out of line, check its working.

x	−2	−1	0	1	2	3	4
y	5	4	3	2	1	0	−1

Check. All the plotted points lie on one straight line. A point taken from the line, such as one with a different x, also satisfies the equation. ✓

Watch out. Use a ruler and extend the line across the whole grid. If the points do not line up, one y value is wrong.

Done

FDraw the graph of y = 3x − 2 for values of x from −1 to 3.

Method. Draw a table of values, plot the points, then join them with a straight line.

Make a table of values. Two points are enough to draw a straight line, but three or more let you spot a mistake: all the points must line up.
Work out y for each x. Substitute each x into the equation.
x = −1: y = 3 × (−1) − 2 = −3 − 2 = −5
x = 0: y = 3 × 0 − 2 = 0 − 2 = −2
x = 1: y = 3 × 1 − 2 = 3 − 2 = 1
x = 2: y = 3 × 2 − 2 = 6 − 2 = 4
x = 3: y = 3 × 3 − 2 = 9 − 2 = 7
x −1 0 1 2 3
y −5 −2 1 4 7
Plot the points. Each pair is a coordinate: across first, then up or down.
(−1, −5) (0, −2) (1, 1) (2, 4) (3, 7)
Join them with a ruler and extend the line. The graph of a linear equation is a straight line, so the points must line up. If one is out of line, check its working.

x	−1	0	1	2	3
y	−5	−2	1	4	7

Check. All the plotted points lie on one straight line. A point taken from the line, such as one with a different x, also satisfies the equation. ✓

Watch out. Use a ruler and extend the line across the whole grid. If the points do not line up, one y value is wrong.

Done

FDraw the graph of y = ½x + 2 for values of x from −4 to 4. (Use even values of x.)

Method. Draw a table of values, plot the points, then join them with a straight line.

Make a table of values. Two points are enough to draw a straight line, but three or more let you spot a mistake: all the points must line up.
Work out y for each x. Substitute each x into the equation.
x = −4: y = 12 × (−4) + 2 = −2 + 2 = 0
x = −2: y = 12 × (−2) + 2 = −1 + 2 = 1
x = 0: y = 12 × 0 + 2 = 0 + 2 = 2
x = 2: y = 12 × 2 + 2 = 1 + 2 = 3
x = 4: y = 12 × 4 + 2 = 2 + 2 = 4
x −4 −2 0 2 4
y 0 1 2 3 4
Plot the points. Each pair is a coordinate: across first, then up or down.
(−4, 0) (−2, 1) (0, 2) (2, 3) (4, 4)
Join them with a ruler and extend the line. The graph of a linear equation is a straight line, so the points must line up. If one is out of line, check its working.

x	−4	−2	0	2	4
y	0	1	2	3	4

Check. All the plotted points lie on one straight line. A point taken from the line, such as one with a different x, also satisfies the equation. ✓

Watch out. Use a ruler and extend the line across the whole grid. If the points do not line up, one y value is wrong.

Done

FIs each point on the line y = 2x − 3? Write yes or no and show your working.

(a)(4, 5)

(b)(−1, −5)

(c)(0, 3)

(d)(3, 2)

Done

PSA line has equation y = 3x + 2. (a) Find the value of y when x = 4. (b) Find the value of x when y = 11.

Done

PSThe points (0, −1), (1, 1) and (2, 3) are on a straight line. Which equation is the equation of the line? A: y = x − 1 B: y = 2x − 1 C: y = 2x + 1

Done

RSam says the line y = 3 is a vertical line. Is he right? Give a reason.

Done

PSThe lines x = −2, x = 3, y = −1 and y = 3 form a rectangle. Work out its perimeter and its area.

Done

RHere is a table of values for a straight line.

x	0	1	2	3
y	1	4	7	11

One of the y values is wrong. Which one, and what should it be? Write the equation of the line.

Done

PSThe lines y = x, x = 4 and the x-axis form a triangle. Work out the area of the triangle.

Done

12.4

Gradient

Edexcel 1MA1: A10, A9
Grade 3-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: GradientComing soon

The gradient of a line tells you how steep it is. A steep hill has a big gradient; a gentle slope has a small one; flat ground has a gradient of zero. To measure it, compare how far the line goes up with how far it goes across:

gradient = rise ÷ run = (change in y) ÷ (change in x)

The rise is the vertical change and the run is the horizontal change. Always move from left to right, so the run is positive. Then the rise tells you the direction: if the line goes up the rise is positive, and if it goes down the rise is negative.

Lines that slope up have a positive gradient, lines that slope down have a negative gradient, and a flat line has a gradient of 0. The bigger the number (ignoring the sign), the steeper the line.

Finding the gradient from a graph

Choose two points on the line that sit exactly on grid corners, as far apart as you sensibly can. Draw a right-angled triangle under the line, count the squares for the run and the rise, and divide. The picture shows a line with a rise of 6 for a run of 3, so its gradient is 6 ÷ 3 = 2: it goes up 2 for every 1 step across.

The gradient triangle on the line y = 2x − 1: rise ÷ run = 6 ÷ 3 = 2.

Finding the gradient from two points

You do not need a picture. If the points are (x₁, y₁) and (x₂, y₂), the rise is y₂ − y₁ and the run is x₂ − x₁:

gradient = (y₂ − y₁) ÷ (x₂ − x₁)

It does not matter which point you call point 1, as long as you subtract in the same order on the top and on the bottom. Simplify the fraction at the end.

Special cases

A horizontal line has a rise of 0, so its gradient is 0. A vertical line has a run of 0, and you cannot divide by 0, so a vertical line has no gradient (we say it is undefined). Gradients are often written as fractions, for example 2/3 or −3/4. Road signs show the gradient as a percentage: a gradient of 0.08 is “8%”.

Key ideas

Gradient = rise ÷ run = (change in y) ÷ (change in x). Read the line left to right.
Up from left to right: positive. Down from left to right: negative. Horizontal: 0. Vertical: undefined.
From two points: (y₂ − y₁) ÷ (x₂ − x₁), subtracting in the same order top and bottom.
The steeper the line, the bigger the gradient (ignoring the sign).
Simplify gradient fractions, and give the sign.

Worked example 12.4.1

Work out the gradient of the line y = 2x − 1 shown on the grid. (The equation is not needed: use the graph.)

Key words. The gradient of a line is how steep it is: rise (the change up or down) divided by run (the change across, to the right). A line sloping up from left to right has a positive gradient; sloping down has a negative gradient.

Method. Gradient = rise ÷ run. Draw a triangle on the line and count the squares up (or down) and across.

Choose two points on the line that are exactly on grid corners. Points that sit on crossing grid lines are easy to read accurately. Choose them far apart for the best accuracy.
(−1, −3) and (3, 5)
Draw a right-angled triangle under the line and count the run. The run is the distance across, moving to the right.
run = 3 − (−1) = 4
Count the rise. The rise is the distance up (positive) or down (negative) while moving right along the line.
rise = 5 − (−3) = 8
Mark the triangle on the graph.
Divide the rise by the run. Gradient = rise ÷ run, simplified.
gradient = 8 ÷ 4 = 2

Check. The line goes up from left to right, so the gradient should be positive. Also, for every 1 step across the line goes up 2. ✓

Watch out. If the line goes down as you move right, the rise is negative. Always read the line from left to right.

Worked example 12.4.2

Work out the gradient of this line.

Method. Gradient = rise ÷ run. Draw a triangle on the line and count the squares up (or down) and across.

Choose two points on the line that are exactly on grid corners. Points that sit on crossing grid lines are easy to read accurately. Choose them far apart for the best accuracy.
(−2, 7) and (4, −5)
Draw a right-angled triangle under the line and count the run. The run is the distance across, moving to the right.
run = 4 − (−2) = 6
Count the rise. The rise is the distance up (positive) or down (negative) while moving right along the line.
rise = −5 − 7 = -12
Mark the triangle on the graph.
Divide the rise by the run. Gradient = rise ÷ run, simplified.
gradient = -12 ÷ 6 = −2

Check. The line goes down from left to right, so the gradient should be negative. Also, for every 1 step across the line goes down 2. ✓

Watch out. If the line goes down as you move right, the rise is negative. Always read the line from left to right.

Worked example 12.4.3

Work out the gradient of the line through A(−3, −4) and B(5, 0).

Method. Gradient = rise ÷ run = (change in y) ÷ (change in x) = (y₂ − y₁) ÷ (x₂ − x₁).

Label the points. Call the first point (A) (x₁, y₁) and the second (B) (x₂, y₂). It does not matter which is first, but you must be consistent.
A = (−3, −4), B = (5, 0)
Find the rise: the change in y. Rise is how far the line goes up (positive) or down (negative). Subtract the y-coordinates, second minus first.
rise = y₂ − y₁ = 0 − (−4) = 4
Find the run: the change in x. Run is how far across to the right. Subtract the x-coordinates in the same order as the y-coordinates.
run = x₂ − x₁ = 5 − (−3) = 8
Sketch the rise and run. The dashed triangle shows the run (across) and the rise (up or down) between the two points.
Divide the rise by the run and simplify. Gradient = rise ÷ run. Cancel the fraction if you can. A negative divided by a positive (or the reverse) is negative.
gradient = rise ÷ run = 4 ÷ 8 = 12

Check. Swap the points: (−4 − 0) ÷ (−3 − 5) = −4 ÷ (−8) = 12, the same. The gradient is positive: the line goes up from left to right. ✓

Watch out. Subtract in the same order on the top and the bottom. Mixing the order (y₂ − y₁ over x₁ − x₂) gives the wrong sign.

Worked example 12.4.4

A road climbs 15 m over a horizontal distance of 200 m. Work out the gradient of the road, as a fraction in its simplest form and as a percentage.

Method. Gradient = rise ÷ run. Both must be in the same units, so convert the rise to metres first.

Write the rise and the run in the same units. A fraction compares the vertical change with the horizontal change. Here both are in metres.
rise = 15 m, run = 200 m
Write the gradient as a fraction. Gradient = rise ÷ run.
15 ÷ 200 = 15200
Simplify the fraction. Divide the top and the bottom by their highest common factor, 5.
15200 = 340
Convert to a percentage if needed. A road gradient is often given as a percentage: multiply the decimal by 100.
3 ÷ 40 = 0.075, so 0.075 × 100 = 7.5%

Check. Going 200 m along the flat, the road rises 7.5 m for every 100 m, which is 15 m in 200 m. ✓

Watch out. Rise and run must use the same units: do not divide metres by kilometres.

Watch out

Rise over run, never run over rise. The gradient is (up) ÷ (across). Also take care with signs: a line that goes down as you move right has a negative gradient. And always subtract in the same order on the top and the bottom of the fraction: y₂ − y₁ goes with x₂ − x₁.

12.5

The equation y = mx + c

Edexcel 1MA1: A9, A10
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Every straight line (that is not vertical) has an equation of the form y = mx + c. The two letters m and c are numbers that describe the line:

m is the gradient: the number multiplying x. It tells you how steep the line is and which way it slopes.
c is the y-intercept: the number on its own. It is the value of y where the line crosses the y-axis, because on the y-axis x = 0 and so y = m × 0 + c = c.

For example, y = 3x + 2 has gradient 3 and crosses the y-axis at (0, 2). The line y = −x + 5 has gradient −1 (a minus sign on its own means −1) and y-intercept 5. The line y = 4x has no c showing, so c = 0 and it passes through the origin. Watch the order: in y = 5 − 3x the gradient is −3 and the intercept is 5, because each sign belongs to the number after it.

The number c is where each line crosses the y-axis (dots). Lines with the same m have the same slope: y = x + 1 and y = x − 2 are parallel.

Sketching a line using m and c

You can sketch a line without a table of values. Step 1: mark the y-intercept (0, c). Step 2: use the gradient as “across then up (or down)” to find another point: for a gradient of 2 = 2/1 go 1 across and 2 up; for −3/4 go 4 across and 3 down. Step 3: join the points with a ruler.

Lines that are not written as y = …

If the equation is not in the form y = mx + c, rearrange it first so that y is on its own. For example 2y = 6x + 4: divide every term by 2 to get y = 3x + 2. And y − 2x = 3: add 2x to both sides to get y = 2x + 3. You met rearranging in Chapters 8 and 9.

Finding the axis crossing points

A line crosses the y-axis where x = 0 and the x-axis where y = 0. For equations such as 2x + 3y = 12 this is a quick way to sketch: put x = 0 to get 3y = 12 so y = 4, giving (0, 4); put y = 0 to get 2x = 12 so x = 6, giving (6, 0). Join the two points.

Higher: you will be asked to rearrange equations such as 3x + 4y = 12 into the form y = mx + c to find the gradient and intercept. The steps are: subtract 3x from both sides to get 4y = −3x + 12, then divide every term by 4 to get y = −34x + 3. The gradient is −34 and the intercept is 3.

Key ideas

y = mx + c: m = gradient (number multiplying x), c = y-intercept (the point (0, c)).
Keep the sign with the number: in y = 5 − 3x, m = −3 and c = 5.
Sketch: plot (0, c), then use the gradient (across, then up or down), then join with a ruler.
If y is not on its own, rearrange first. Divide every term.
The y-axis crossing: x = 0. The x-axis crossing: y = 0.

Worked example 12.5.1

Write down the gradient and the y-intercept of the line y = −2x + 7.

Method. Write the equation as y = mx + c. Then m is the gradient and c is the y-intercept.

Compare with y = mx + c. The equation is already in the form y = mx + c, so we can read it off.
y = −2x + 7
The gradient m is the number multiplying x. Include its sign. If there is no number, x means 1x and m = 1; if there is only a minus sign, m = −1.
m = −2
The y-intercept c is the number on its own. It is where the line crosses the y-axis. If there is no number, c = 0 and the line goes through the origin.
c = 7, so the line crosses the y-axis at (0, 7)

Check. Put x = 0 into the equation: y = 7, which is the y-intercept. Put x = 1: y = 5, which is −2 more than 7, so the gradient is −2. ✓

Watch out. Keep the sign with the number. In y = 5 − 3x the gradient is −3 (not 3) and the intercept is 5 (not −3).

Worked example 12.5.2

Write down the gradient and the y-intercept of the line y = 5 − 3x.

Method. Write the equation as y = mx + c. Then m is the gradient and c is the y-intercept.

Write the equation in the form y = mx + c. Put the x term first and the plain number last, keeping the sign in front of each term.
y = 5 − 3x = y = −3x + 5
The gradient m is the number multiplying x. Include its sign. If there is no number, x means 1x and m = 1; if there is only a minus sign, m = −1.
m = −3
The y-intercept c is the number on its own. It is where the line crosses the y-axis. If there is no number, c = 0 and the line goes through the origin.
c = 5, so the line crosses the y-axis at (0, 5)

Check. Put x = 0 into the equation: y = 5, which is the y-intercept. Put x = 1: y = 2, which is −3 more than 5, so the gradient is −3. ✓

Watch out. Keep the sign with the number. In y = 5 − 3x the gradient is −3 (not 3) and the intercept is 5 (not −3).

Worked example 12.5.3

Sketch the graph of y = 12x − 2 using the gradient and the y-intercept.

Method. Start at the y-intercept, then use the gradient (rise ÷ run) to find more points. No table of values is needed.

Find the y-intercept and mark it. The number on its own is c. The line crosses the y-axis at (0, c), so this is the first point.
c = −2: mark (0, −2)
Use the gradient to find a second point. The gradient is 12, which means: go 2 across and 1 up (rise ÷ run = 1/2).
from (0, −2) go 2 right and 1 up to (2, −1)
Go the other way to get a third point as a check. Going left by the same amount means doing the opposite of the rise too.
from (0, −2) go 2 left and 1 down to (−2, −3)
Draw a straight line through the points with a ruler. Extend the line across the grid. Label it with its equation.

Check. Pick the point with x = 2: y = 12 × 2 − 2 = −1, which is on the sketch. ✓

Watch out. For a negative gradient go down as you go across (or up if you go left).

Worked example 12.5.4

Write 2y = 6x + 4 in the form y = mx + c, and write down the gradient and the y-intercept.

Method. Divide every term by the number in front of y to get the form y = mx + c.

The equation has a number multiplying y. We want y on its own (coefficient 1), so divide both sides by that number.
Start: 2y = 6x + 4
Divide every term by 2. Every term on the right must be divided too, not just the first.
y = 6x ÷ 2 + 4 ÷ 2
y = 3x + 2
Read off the gradient and the y-intercept. Compare with y = mx + c.
m = 3, c = 2

Check. Test x = 2: the original gives 2y = 16, so y = 8. The new equation gives y = 3 × 2 + 2 = 8. ✓

Watch out. Divide every term by the number in front of y, including the plain number.

Worked example 12.5.5

Find the gradient and the y-intercept of the line 3x + 4y = 12.

▶ Watch this solution as a video

Method. Rearrange to make y the subject. Whatever you do to one side you must do to the other side, and divide every term.

Get the y term on its own on one side. We want the equation to start y = ..., so move everything except the y term to the other side using the opposite operation, on both sides.
Start: 3x + 4y = 12
Subtract 3x from both sides: 4y = −3x + 12
Divide every term by the number in front of y. y must have a coefficient of 1. Divide every term (both of them) by 4.
Divide every term on both sides by 4: y = −34x + 3
Read off the gradient and the intercept. Compare with y = mx + c: the number in front of x is m and the number on its own is c.
m = −34, c = 3

Check. Test x = 3. Original: 3 × 3 + 4 × y = 12 gives y = 0.75. New form: y = (−34) × 3 + 3 = 0.75. They match. ✓

Watch out. When you divide by a number, divide every term, including the x term. Moving a term across changes its sign.

Worked example 12.5.6

Sketch the line 2x + 3y = 12 by finding where it crosses the axes.

Key words. The x-intercept is where the line crosses the x-axis (y = 0). The y-intercept is where it crosses the y-axis (x = 0).

Method. Find where the line crosses each axis by putting x = 0 and then y = 0. Join the two points.

Find where the line crosses the y-axis: put x = 0. Every point on the y-axis has x = 0. The x term disappears.
3y = 12 → y = 12 ÷ 3 = 4
The line crosses the y-axis at (0, 4).
Find where the line crosses the x-axis: put y = 0. Every point on the x-axis has y = 0. The y term disappears.
2x = 12 → x = 12 ÷ 2 = 6
The line crosses the x-axis at (6, 0).
Plot the two points and join them with a ruler. Two points are enough for a straight line. Extend the line beyond them.

Check. Test a third point, such as x = 8: y = (12 − 2 × 8) ÷ 3 = −43, and the point is on the sketch. ✓

Watch out. For the y-intercept the x value is 0; for the x-intercept the y value is 0. Do not mix them up.

Watch out

Signs and order. The sign belongs to the term after it. In y = 5 − 3x, the gradient is −3 (not 3), and the y-intercept is 5 (not −3). And when you rearrange, divide every term: 2y = 6x + 4 gives y = 3x + 2, not y = 3x + 4.

12.6

Finding the equation of a straight line

Edexcel 1MA1: A9, A10
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

To find the equation of a line you need m and c for y = mx + c. There are three common situations.

1. From a graph

Read c where the line crosses the y-axis. Find m with a gradient triangle (rise ÷ run) using two points on grid corners. Then write y = mx + c. If the line is horizontal its equation is y = a number; if it is vertical its equation is x = a number.

2. From the gradient and one point

Write y = mx + c and put in the gradient. The point is on the line, so its coordinates make the equation true. Substitute them for x and y and solve the little equation for c. For example, gradient 3 through (2, 5): 5 = 3 × 2 + c, so 5 = 6 + c and c = −1. The equation is y = 3x − 1.

3. From two points

First work out the gradient using (y₂ − y₁) ÷ (x₂ − x₁). Then you have a gradient and a point (use either), so continue as in case 2. Always check with the other point.

Reading an equation from a graph: c = 3 where the line crosses the y-axis. The gradient triangle has a run of 6 and a rise of −3, so m = −3 ÷ 6 = −½.

Why does substituting a point work? Because every point on the line obeys the equation. Once we know m, the only unknown left is c, and one point is enough to find it. If you are asked for the equation of a line that is parallel to another or that passes through a given point, you are using exactly these methods (see Section 12.7).

Some graphs use a different scale on the two axes, or the line does not cross at an easy point. Then use two points that you can read exactly, as far apart as possible, and use the two-points method.

Higher: another way to write the same equation is y − y₁ = m(x − x₁), where (x₁, y₁) is a point on the line. It is useful because you do not have to find c separately: for gradient 3 through (2, 5), y − 5 = 3(x − 2), which expands and rearranges to y = 3x − 1. You can also find m and c by writing one equation for each point and solving them together: this method returns in Chapter 14.

Key ideas

From a graph: c = y-intercept, m = rise ÷ run from a triangle, then y = mx + c.
From m and a point: substitute the point into y = mx + c and solve for c.
From two points: find m first, then substitute one point to find c; check with the other point.
Horizontal line: y = number. Vertical line: x = number.
Always write the final answer as an equation y = …

Worked example 12.6.1

Find the equation of the line shown.

Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.

Find where the line crosses the y-axis. That value is c, the y-intercept.
The line crosses the y-axis at (0, 1), so c = 1.
Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
Work out the gradient. Gradient = rise ÷ run.
rise = 10, run = 5
m = 10 ÷ 5 = 2
Write the equation in the form y = mx + c. Put m and c into the formula.
y = 2x + 1

Check. Test a point on the line, (−3, −5): y = 2 × (−3) + 1 = −5, which matches. ✓

Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.

Worked example 12.6.2

Find the equation of the line shown.

Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.

Find where the line crosses the y-axis. That value is c, the y-intercept.
The line crosses the y-axis at (0, 3), so c = 3.
Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
Work out the gradient. Gradient = rise ÷ run.
rise = -3, run = 6
m = -3 ÷ 6 = −12
Write the equation in the form y = mx + c. Put m and c into the formula.
y = −12x + 3

Check. Test a point on the line, (−4, 5): y = (−12) × (−4) + 3 = 5, which matches. ✓

Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.

Worked example 12.6.3

Find the equation of the line with gradient 3 that passes through the point (2, 5).

Method. Use y = mx + c. Put in the gradient, substitute the coordinates of the point, then solve for c.

Write the general equation of a straight line. Every straight line (not vertical) has the form y = mx + c.
y = mx + c
Put in the gradient. We are told the gradient is m = 3.
y = 3x + c
Substitute the point P (2, 5) into the equation. The point is on the line, so when x and y take its values, the equation is true.
5 = 3 × 2 + c
5 = 6 + c
Solve for c. Use the inverse operation on both sides.
c = 5 − 6 = −1
Write the equation.
y = 3x − 1

Check. Substitute the point back: y = 3 × 2 − 1 = 5, which is 5. ✓

Watch out. Substitute x into the x position and y into the y position. Do not swap them.

Worked example 12.6.4

Find the equation of the line through A(1, 5) and B(4, 11).

Method. Find the gradient from the two points, then substitute one point into y = mx + c to find c.

Find the gradient first. Use gradient = change in y ÷ change in x, subtracting both in the same order.
m = (11 − 5) ÷ (4 − 1) = 6 ÷ 3 = 2
Put the gradient into y = mx + c. Now the only thing we do not know is c.
y = 2x + c
Substitute one of the points to find c. Use A (1, 5). The point is on the line, so its coordinates must fit the equation. Replace x and y by the point’s numbers.
5 = 2 × 1 + c
5 = 2 + c
Solve for c. Take 2 from both sides (the opposite of adding 2).
c = 5 − 2 = 3
Write the full equation. Put m and c into y = mx + c.
y = 2x + 3

Check. Test the other point B (4, 11): y = 2 × 4 + 3 = 11, which is 11. ✓

Watch out. Check with the second point. If it does not fit, recheck the gradient: subtract in the same order on the top and the bottom.

Worked example 12.6.5

Find the equation of the line through A(−2, 7) and B(4, −2).

▶ Watch this solution as a video

Method. Find the gradient from the two points, then substitute one point into y = mx + c to find c.

Find the gradient first. Use gradient = change in y ÷ change in x, subtracting both in the same order.
m = (−2 − 7) ÷ (4 − (−2)) = −9 ÷ 6 = −32
Put the gradient into y = mx + c. Now the only thing we do not know is c.
y = −32x + c
Substitute one of the points to find c. Use A (−2, 7). The point is on the line, so its coordinates must fit the equation. Replace x and y by the point’s numbers.
7 = (−32) × (−2) + c
7 = 3 + c
Solve for c. Take 3 from both sides (the opposite of adding 3).
c = 7 − 3 = 4
Write the full equation. Put m and c into y = mx + c.
y = −32x + 4

Check. Test the other point B (4, −2): y = (−32) × 4 + 4 = −2, which is −2. ✓

Watch out. Check with the second point. If it does not fit, recheck the gradient: subtract in the same order on the top and the bottom.

Watch out

Do not use the y-coordinate of a point as c. Only the point (0, c) tells you c directly. For any other point you must substitute. And take care with negative gradients: reading rise ÷ run on a line that goes down gives a negative rise.

12.7

Parallel and perpendicular lines

Edexcel 1MA1: A9
Grade 4-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Parallel lines go in the same direction and never meet, like the two rails of a straight railway track. They have exactly the same steepness, so parallel lines have equal gradients. In y = mx + c form, the lines y = 2x + 1, y = 2x − 3 and y = 2x are all parallel because every one has m = 2. Only the y-intercept (c) is different, which just slides the line up or down.

If an equation is not in the form y = mx + c, rearrange it first before comparing gradients. For example 2y = 4x + 6 becomes y = 2x + 3, so it is parallel to y = 2x + 1.

Parallel lines: all three have gradient 2, so they slope the same way and never meet.

Finding a line parallel to another

To find the equation of a line parallel to y = 2x + 1 that passes through (3, 4): the gradient must be 2, so write y = 2x + c, substitute the point to get 4 = 6 + c, so c = −2. The line is y = 2x − 2. Keep the gradient, find a new c.

Perpendicular lines (Higher)

Two lines are perpendicular if they meet at a right angle (90°). Their gradients are linked: multiply the two gradients and you get −1. Equivalently, the perpendicular gradient is the negative reciprocal: flip the fraction upside down and change the sign.

Examples: a line with gradient 2 (which is 2/1) has a perpendicular line with gradient −1/2. A line with gradient −3/4 has a perpendicular line with gradient 4/3. Why does this work? Take a line with gradient 2: it goes 1 across and 2 up. Turn it a quarter turn anticlockwise: it now goes 2 across the opposite way and 1 up, which is “across −2, up 1”, a gradient of 1 ÷ (−2) = −1/2. A quarter turn swaps the rise and the run and changes one sign. Check: 2 × (−1/2) = −1.

The lines y = 2x − 1 and y = −½x + 4 meet at a right angle at (2, 3). The gradients 2 and −½ multiply to −1.

Equation of a perpendicular line (Higher)

To find the line perpendicular to y = 2x + 1 through (4, 1): the perpendicular gradient is −½, so y = −½x + c. Substitute (4, 1): 1 = −2 + c, so c = 3. A perpendicular bisector of a line segment passes through its midpoint and is perpendicular to the segment.

Key ideas

Parallel lines have equal gradients. Rearrange to y = mx + c to compare.
A parallel line through a point: keep m, substitute the point to find the new c.
Higher: perpendicular gradients multiply to −1: the perpendicular gradient is the negative reciprocal (flip and change sign).
Higher: perpendicular bisector = through the midpoint, with the perpendicular gradient.

Worked example 12.7.1

Are the lines y = 3x − 4 and 3y = 9x + 2 parallel?

Method. Rearrange the second equation into the form y = mx + c, then compare the gradients.

Find the gradient of the first line. The equation is already y = mx + c.
y = 3x − 4 has m = 3
Rearrange the second equation. Divide every term by 3 so that y is on its own.
3y = 9x + 2 → y = 3x + 23
Compare the gradients. Parallel lines have the same gradient.
3 = 3
Write the conclusion.
The lines are parallel.

Check. Both lines rise 3 for every 1 across. ✓

Watch out. You must rearrange first. In 3y = 9x + 2 the gradient is not 9.

Worked example 12.7.2

Find the equation of the line that is parallel to y = 2x + 1 and passes through the point (3, 4).

Method. Parallel lines have the same gradient. Keep m, then use the point to find the new c.

Read the gradient of the given line. The equation is in the form y = mx + c, so m is the number in front of x.
y = 2x + 1 has gradient m = 2
Use the same gradient. Parallel lines have equal gradients: they have the same steepness so they never meet.
y = 2x + c
Substitute the point (3, 4) to find c. The new line must go through this point.
4 = 2 × 3 + c
4 = 6 + c
c = 4 − 6 = −2
Write the equation.
y = 2x − 2

Check. The gradient 2 matches the given line, and the point (3, 4) fits: 2 × 3 − 2 = 4. ✓

Watch out. Only the gradient stays the same. If you keep c as well you get the same line, not a parallel one.

Worked example 12.7.3

Find the equation of the line that is parallel to y = −12x + 3 and passes through (4, 5).

Method. Parallel lines have the same gradient. Keep m, then use the point to find the new c.

Read the gradient of the given line. The equation is in the form y = mx + c, so m is the number in front of x.
y = −12x + 3 has gradient m = −12
Use the same gradient. Parallel lines have equal gradients: they have the same steepness so they never meet.
y = −12x + c
Substitute the point (4, 5) to find c. The new line must go through this point.
5 = (−12) × 4 + c
5 = −2 + c
c = 5 − (−2) = 7
Write the equation.
y = −12x + 7

Check. The gradient −12 matches the given line, and the point (4, 5) fits: (−12) × 4 + 7 = 5. ✓

Watch out. Only the gradient stays the same. If you keep c as well you get the same line, not a parallel one.

Worked example 12.7.4

Find the gradient of a line that is perpendicular to a line with gradient 23.

▶ Watch this solution as a video

Key words. Two lines are perpendicular if they meet at a right angle (90°). The reciprocal of a number is 1 divided by it; the negative reciprocal also changes the sign.

Method. Perpendicular lines meet at a right angle. Their gradients multiply to give −1, so the perpendicular gradient is the negative reciprocal: flip the fraction and change the sign.

Write the gradient as a fraction. A whole number is a fraction over 1. This makes it easy to flip.
m = 23
Flip the fraction upside down. The perpendicular gradient uses the reciprocal: the top and the bottom swap places.
23 flips to 32
Change the sign. The perpendicular gradient is the negative reciprocal: positive becomes negative, negative becomes positive.
perpendicular gradient = −32

Check. Multiply the gradients: 23 × (−32) = −1. ✓

Watch out. You need both changes: flip and change sign. The reciprocal of 2 is ½, but the perpendicular gradient is −½.

Worked example 12.7.5

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

▶ Watch this solution as a video

Method. Perpendicular lines have gradients that multiply to −1. Find the negative reciprocal of the gradient, then use the point to find c.

Read the gradient of the given line. m is the number in front of x when the equation is y = mx + c.
y = 2x + 1 has gradient 2
Find the perpendicular gradient. Flip the fraction and change the sign (the negative reciprocal). Check: the two gradients must multiply to give −1.
2 → −12
Check: 2 × (−12) = −1
Put the new gradient in y = mx + c. Only c is unknown now.
y = −12x + c
Substitute the point (4, 1) to find c. The new line must pass through this point.
1 = (−12) × 4 + c
1 = −2 + c
c = 1 − (−2) = 3
Write the equation.
y = −12x + 3

Check. The point fits: (−12) × 4 + 3 = 1 = 1. And 2 × (−12) = −1. ✓

Watch out. Flip and change the sign. Then substitute the point into the new equation, not the old one.

Worked example 12.7.6

Find the equation of the perpendicular bisector of the line segment from A(0, 1) to B(4, 3).

▶ Watch this solution as a video

Key words. A bisector cuts something exactly in half. A perpendicular bisector of a line segment cuts it in half at 90°.

Method. A perpendicular bisector cuts a line segment in half at right angles. It goes through the midpoint and has the negative reciprocal gradient.

Find the midpoint of AB. A bisector cuts the line in half, so it passes through the midpoint.
M = ((0 + 4) ÷ 2, (1 + 3) ÷ 2) = (2, 2)
Find the gradient of AB. Rise ÷ run.
m = (3 − 1) ÷ (4 − 0) = 2 ÷ 4 = 12
Find the perpendicular gradient. “Perpendicular” means at right angles to AB, so use the negative reciprocal.
12 → −2
Substitute the midpoint into y = mx + c. The bisector goes through the midpoint M.
2 = (−2) × 2 + c
2 = −4 + c
c = 2 − (−4) = 6
Write the equation.
y = −2x + 6

Check. 12 × (−2) = −1, so it is perpendicular; and the midpoint fits: (−2) × 2 + 6 = 2 = 2. ✓

Watch out. Use the midpoint for the point, not one of the end points.

Watch out

Same gradient, not same c. Parallel lines have equal m; if c is also equal they are the same line. Always rearrange to y = mx + c first: the gradient of 2y = 6x + 1 is 3, not 6. For perpendicular lines you must flip and change the sign.

12.8

Real-life graphs and rates of change

Edexcel 1MA1: A14, R14, A10
Grade 3-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Real-life graphs and rates of changeComing soon

Straight-line graphs are used to describe real situations where one quantity changes steadily with another: converting between units, the cost of a taxi, a phone tariff, water filling a tank. The graph is a quick way to read off values, and the equation y = mx + c lets you calculate them exactly.

What the gradient and intercept mean

In a real-life graph the gradient is the rate of change: how much the quantity on the vertical axis changes for each 1 unit on the horizontal axis. Its units are “vertical units per horizontal unit”. In a conversion graph from miles to km the gradient is 1.6 km per mile. In a taxi fare graph it is the cost per mile. A negative gradient means the quantity is decreasing, like water draining from a tank.

The y-intercept is the starting value, when the horizontal value is 0. For a taxi, it is the fixed charge you pay before travelling any distance. For a graph that goes through the origin, the starting value is 0: this is a graph of direct proportion, y = kx, where doubling x doubles y.

Converting miles to kilometres. To convert 30 miles, go up from 30 on the horizontal axis to the line, then across: 48 km. The gradient is 1.6 km per mile.

Reading values from a graph

To find a value, go from the known value on one axis to the line, then to the other axis. Use a ruler and dotted lines. First check the scale: work out what each small square is worth on each axis. To find an exact value for an equation such as C = 2n + 3, substitute into the formula instead.

Comparing two graphs

When two companies charge differently, draw both lines on the same axes. The point where the lines cross is where the two costs are equal. Before that point one company is cheaper, after it the other is. A line with a smaller starting charge but a bigger gradient is cheaper for small amounts and more expensive for large amounts.

Plan A costs £10 a month plus £2 per GB; Plan B costs £4 a month plus £3 per GB. The lines cross at 6 GB where both cost £22.

Journey graphs

In a distance–time graph the gradient is the speed (distance ÷ time). A steeper line means faster travel; a horizontal line means the object is not moving (the distance is not changing); a line going back down means returning towards the start. You will meet these again, with curves, in Chapter 13, and speed in Chapter 16.

A cycle ride: out for 1 hour (12 km), a rest of half an hour, then back home. The gradient of each section is the speed.

Direct and inverse proportion graphs

If y is directly proportional to x, the graph is a straight line through the origin (graph A below). A straight line that does not go through the origin (B) is a linear relationship but not direct proportion. If y is inversely proportional to x, the graph is a curve that falls as x rises (C): see Chapters 13 and 16.

A: direct proportion (straight line through the origin). B: straight line, not proportional. C: inverse proportion (a curve).

Key ideas

The gradient of a real-life graph is the rate of change: units are (vertical units) per (horizontal unit).
The y-intercept is the starting value (when the horizontal value is 0), for example a fixed charge.
Direct proportion: a straight line through the origin, y = kx. Inverse proportion: a curve.
Where two lines cross, the two quantities are equal.
On a distance–time graph, the gradient is the speed and a horizontal line means stopped.
Read the scale of each axis carefully before you read values.

Worked example 12.8.1

The graph converts miles to kilometres. Use the graph to convert 30 miles to kilometres.

Key words. To read off a value from a graph, draw guide lines from one axis to the line and then across to the other axis.

Method. Read across from the graph: go up from the horizontal value to the line, then across to the vertical axis.

Find 30 miles on the horizontal axis. The horizontal axis shows the distance in miles.
distance in miles: 30 miles
Go straight up to the line. This finds the point on the graph that has this horizontal value. Use a ruler and a dotted line.
Go across to the vertical axis and read the value. The vertical axis shows the distance in km. The dotted line meets it at the answer.
distance in km: 48 km

Check. Use the formula k = 1.6m: k = 1.6 × 30 = 48. ✓

Watch out. Read the scale carefully: count how much each small square is worth on each axis.

Worked example 12.8.2

The graph shows the cost, £C, of a taxi journey of n miles. Find a formula for C in terms of n.

Method. Find the y-intercept and the gradient, then write the equation in the form y = mx + c with the letters given in the question.

Find the y-intercept c. Read where the line crosses the vertical axis.
c = 3
Find the gradient m. Use rise ÷ run on two points of the line.
(23 − 3) ÷ 10 = 20 ÷ 10 = 2
Write y = mx + c using the letters from the question. Replace x by n and y by C, since these stand for the real quantities.
C = 2n + 3

Check. Put n = 10 into the formula: 2 × 10 + 3 = 23, which is the height of the line there. ✓

Watch out. Use the letters in the question, not x and y, if the question defines other letters.

Worked example 12.8.3

For the taxi graph above, find the gradient and say what it represents.

Method. The gradient of a real-life graph is rise ÷ run, and it is the rate of change: how much the vertical quantity changes for each 1 on the horizontal axis.

Choose two points on the line. Choose points that are easy to read and far apart.
(0, 3) and (10, 23)
Draw a triangle and work out the rise and the run. Rise = change in the vertical value; run = change in the horizontal value.
rise = 23 − 3 = 20
run = 10 − 0 = 10
Divide the rise by the run. Gradient = rise ÷ run.
gradient = 20 ÷ 10 = 2
Say what the gradient means. The gradient is the change in the cost for each 1 unit increase in the distance. It is a rate of change.
The gradient is £2 per mile.

Check. For each 1 across, the line goes up 2, which matches the gradient. ✓

Watch out. Give the units of the gradient: it is (vertical units) per (horizontal unit), for example £2 per mile.

Worked example 12.8.4

For the taxi graph above, say what the y-intercept means.

Method. The y-intercept is where the line meets the vertical axis. In a real-life graph it is the value at the start (when the horizontal value is 0).

Find where the line crosses the vertical axis. This is where the horizontal value is 0: the very start.
Read the value. The line meets the vertical axis at 3.
cost = £3 when distance = 0 miles
Say what it means. This is the fixed charge: £3 is paid before any miles are travelled.
The y-intercept is 3.

Check. The graph really does start at that height when the horizontal value is 0. ✓

Watch out. Describe the meaning in words: “the fixed charge” or “the starting amount”, not just the number.

Worked example 12.8.5

A taxi fare is given by C = 2n + 3. Work out how far the journey is when the fare is £15.

▶ Watch this solution as a video

Method. Put the known value into the formula and solve the equation for the letter you do not know.

Write the formula and put in the known value. We know the vertical value, so replace the letter for it.
15 = 2n + 3
Subtract 3 from both sides. This removes the fixed amount so only the term with the letter is left.
12 = 2n
Divide both sides by 2. This leaves n on its own.
n = 12 ÷ 2 = 6

Check. Substitute back: 2 × 6 + 3 = 15. ✓

Watch out. Undo the addition first, then the multiplication (reverse order).

Watch out

Read the scale. The small squares on a real-life graph are often worth 2, 5 or 10, not 1. Also give units with the gradient: write “£2 per mile”, not just “2”. And a line that does not start at the origin is not direct proportion even though it is straight.

Exercise 12L Fares, tariffs and comparing lines

Calculator allowed

PSThe graph shows the cost, £C, of a taxi journey of n miles.

(a)Use the graph to find the cost of a 4-mile journey.

(b)Use the graph to find how far you can travel for £15.

(c)What does the y-intercept represent?

(d)Work out the gradient and say what it represents.

(e)Write down a formula for C in terms of n.

Done

PSPlan A costs £10 a month plus £2 for each GB of data. Plan B costs £4 a month plus £3 for each GB of data. Both lines are drawn on the graph.

(a)Which plan is cheaper for 4 GB of data?

(b)Which plan is cheaper for 8 GB of data?

(c)For how much data do the two plans cost the same? What is the cost?

Done

RUse the formulas C = 2g + 10 (Plan A) and C = 3g + 4 (Plan B) to show algebraically that the plans cost the same for 6 GB of data.

Done

RAisha uses a lot of data each month. Which plan (A or B) should she choose? Use the gradients and y-intercepts to explain.

Done

PSA plumber charges a call-out fee of £40 plus £25 for each hour of work. Draw a graph to show the cost, £C, of a job lasting up to 5 hours. Use C = 25h + 40.

Method. Make a table of values from the formula, plot the points and join them.

Write the formula. A fixed call-out charge plus an hourly rate.
C = 25h + 40
Make a table of values. Substitute each h.
h = 0: 25 × 0 + 40 = 40
h = 1: 25 × 1 + 40 = 65
h = 2: 25 × 2 + 40 = 90
h = 3: 25 × 3 + 40 = 115
h = 4: 25 × 4 + 40 = 140
h = 5: 25 × 5 + 40 = 165
h 0 1 2 3 4 5
C (£) 40 65 90 115 140 165
Plot the points and join them. Choose a sensible scale: 1 square for 1 hour and 25 for £25. The line starts at (0, 40) on the vertical axis.

h	0	1	2	3	4	5
C (£)	40	65	90	115	140	165

Check. The points are in a straight line, and the line starts at (0, 40). ✓

Watch out. Start the line on the vertical axis at the call-out charge, not at the origin.

Done

PSA bike hire company charges £5 plus £3 for each hour. (a) Write a formula for the cost, £C, of hiring a bike for h hours. (b) Work out the cost of 4 hours. (c) Hannah pays £29. For how many hours did she hire the bike?

(a)a formula for C

(b)the cost of 4 hours

(c)the number of hours for £29

Done

PSTo hire an electric scooter you pay £1 to unlock it and then 25p for each minute. Write a formula for the cost, £C, of a ride of m minutes, and work out the cost of an 18-minute ride.

Done

PSA water company charges a standing charge of £20 plus £1.80 for each cubic metre (m³) of water used. (a) Write a formula for the cost, £C, of using w m³. (b) Work out the cost of using 150 m³. (c) A household’s bill is £110. How much water did it use?

(a)a formula for C

(b)the cost of 150 m³

(c)the water used for £110

Done

PSCar A leaves a town and travels at a steady 60 km/h. At the same moment car B, which is already 30 km ahead, travels in the same direction at a steady 40 km/h. After how many hours does car A catch up with car B, and how far from the town is that?

Done

PSA candle is lit. After 1 hour it is 20 cm tall and after 3 hours it is 14 cm tall. It burns at a steady rate. Find a formula for its height, h cm, after t hours, and work out when it burns out.

Done

RThe taxi graph does not go through the origin. Explain what this tells you, and what it would mean if the line did go through the origin.

Done

12.9

Where lines cross

Edexcel 1MA1: A9, A10, G11
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

When two lines are drawn on the same grid, the point where they cross is called the point of intersection. It is the only point that lies on both lines, so its coordinates make both equations true. You can find it in three ways.

Draw and read. Draw both lines accurately (from tables of values), then read the coordinates of the crossing point. Check your answer by substituting into both equations.
Lines parallel to the axes. The line x = 3 meets y = 2x − 1 where x = 3; substitute to get y = 5, so the point is (3, 5). The line y = 7 meets it where 7 = 2x − 1, so x = 4. A line crosses the y-axis where x = 0 and the x-axis where y = 0.
Algebra (Higher). At the crossing point the two y values are equal, so set the right-hand sides equal and solve for x, then find y. This leads to simultaneous equations in Chapter 14.

Two lines with the same gradient are parallel: they never cross, so there is no point of intersection. Two different straight lines can cross at most once.

The lines y = 2x − 3 and y = −x + 6 cross at (3, 3).

Solving equations with a graph

A graph can also solve an equation. To solve 2x − 1 = 5, draw the line y = 2x − 1 and the horizontal line y = 5. They cross where 2x − 1 equals 5, and the x-coordinate of that point is the solution, x = 3. You can use this idea whenever you have an equation with the same expression on both sides and a graph of that expression.

Key ideas

The point of intersection lies on both lines, so its coordinates satisfy both equations. Check by substituting into both.
To find where a line meets x = a, put x = a; to find where it meets y = b, put y = b.
To solve an equation graphically, draw y = (left side) and y = (right side); the x of the crossing point is the solution.
Parallel lines (equal gradients) do not intersect.
Higher: set the two expressions for y equal, solve for x, then substitute to find y.

Worked example 12.9.1

The graph shows the lines y = 2x − 3 and y = −x + 6. Write down the coordinates of the point where they cross.

▶ Watch this solution as a video

Key words. The point of intersection of two lines is the point where they cross.

Method. Where two lines cross, the point is on both lines. Read its coordinates from the graph and check them in both equations.

Find the point where the two lines cross. The point of intersection is the only point that is on both lines.
Read its coordinates. Go across from the origin to find x, then up or down to find y.
The lines cross at (3, 3).
Check in both equations. The point must satisfy both equations if it is on both lines.
y = 2x − 3: 2 × 3 − 3 = 3 ✓
y = −x + 6: −3 + 6 = 3 ✓

Check. Both equations give y = 3 when x = 3. ✓

Watch out. Give the answer as a coordinate pair (x, y), not just one number.

Worked example 12.9.2

On the same grid draw the graphs of y = x + 1 and y = 7 − x for x from 0 to 6. Write down the coordinates of the point where the lines cross.

▶ Watch this solution as a video

Method. Draw both lines on one grid, then read the coordinates of the point where they cross.

Make a table of values for y = x + 1. Substitute each x value. Put negatives in brackets.
x = 0: y = 0 + 1 = 1
x = 2: y = 2 + 1 = 3
x = 4: y = 4 + 1 = 5
x = 6: y = 6 + 1 = 7
x 0 2 4 6
y 1 3 5 7
Make a table of values for y = −x + 7. Do the same for the second equation.
x = 0: y = −0 + 7 = 0 + 7 = 7
x = 2: y = −2 + 7 = −2 + 7 = 5
x = 4: y = −4 + 7 = −4 + 7 = 3
x = 6: y = −6 + 7 = −6 + 7 = 1
x 0 2 4 6
y 7 5 3 1
Plot both lines on the same grid. Plot each set of points and join them with a ruler. Label each line.
Read the point where the lines cross. This is the point that is on both lines.
The lines cross at (3, 4).

x	0	2	4	6
y	1	3	5	7

x	0	2	4	6
y	7	5	3	1

Check. Check in both equations: y = x + 1 gives 3 + 1 = 4; y = −x + 7 gives −3 + 7 = 4. ✓

Watch out. Use a sharp pencil and a ruler. A thick or wobbly line makes the crossing point hard to read.

Worked example 12.9.3

Find the coordinates of the point where the line y = 3x − 2 crosses the line y = 7.

▶ Watch this solution as a video

Method. At the point where two lines cross, both equations are true. Use the simple equation to give one coordinate, then substitute to find the other.

The line y = 7 means y is always 7. At the crossing point both equations are true, so y = 7.
y = 7
Substitute into the other equation to find x. Replace y by 7 and solve.
7 = 3x − 2
9 = 3x
x = 9 ÷ 3 = 3
Write the coordinates.
(3, 7)

Check. Put x = 3 in: y = 3 × 3 − 2 = 7. ✓

Watch out. Finish by writing the point as (x, y) with x first.

Worked example 12.9.4

The graph of y = 2x − 1 is drawn on the grid. Use the graph to solve 2x − 1 = 5.

▶ Watch this solution as a video

Key words. To solve an equation graphically means to use a graph to find the value of x that makes it true.

Method. Draw the horizontal line y = (the right-hand number). Where it meets the graph, read off x.

Rewrite the equation as a question about the graph. Solving 2x − 1 = 5 means: for what x does the line y = 2x − 1 have y = 5?
Draw the horizontal line y = 5.
Find where the two lines cross. At the crossing point the two y values are equal, so the x value solves the equation.
Read the x-coordinate of the crossing point. Go straight down from the crossing point to the x-axis.
x = 3

Check. Substitute: 2 × 3 − 1 = 5 = 5. ✓

Watch out. The answer is the x-coordinate, not the y-coordinate of the crossing point.

Worked example 12.9.5

Find the coordinates of the point where the lines y = 2x + 1 and y = −x + 7 cross.

▶ Watch this solution as a video

Key words. The point of intersection is where two lines cross. Its coordinates fit both equations.

Method. At the point of intersection both equations are true, so the two expressions for y are equal. Solve for x, then find y.

Both equations give y, so set the right-hand sides equal. At the point where the lines cross, x and y are the same for both lines.
2x + 1 = −x + 7
Get the x terms on one side. Collect the x terms on one side and the numbers on the other, doing the same to both sides.
3x = 6 (the x terms: 2x − (−1)x = 3x; the numbers: 7 − 1 = 6)
Solve for x. Divide both sides by 3.
x = 6 ÷ 3 = 2
Find y by substituting x into either equation. Use the easier one.
y = 2 × 2 + 1 = 5
Write the point of intersection.
(2, 5)

Check. Put x = 2 into the other equation: y = −2 + 7 = 5. Both give 5. ✓

Watch out. Remember to find y as well: the answer is a coordinate pair, not just a value of x.

Watch out

Give a coordinate pair. The intersection is a point, so write (x, y), not just x. And always check in both equations: if only one works, the crossing point has been misread. When solving an equation graphically, the answer is the x-coordinate of the crossing point, not the y-coordinate.

Exercise 12M Intersections of lines

Non-calculator

FThe graph shows the lines y = x + 1 and y = −x + 5. Write down the coordinates of the point where they cross.

Done

FThe graph shows the lines y = 2x − 3 and y = −x + 6. Write down the coordinates of the point where they cross.

Done

FThe graph shows the lines y = 12x + 1 and y = −x + 4. Write down the coordinates of the point where they cross.

Done

FOn the same grid, draw the graphs of y = x + 2 and y = 4 − x for x from −1 to 4. Write down the coordinates of the point where they cross.

Method. Draw both lines on one grid, then read the coordinates of the point where they cross.

Make a table of values for y = x + 2. Substitute each x value. Put negatives in brackets.
x = −1: y = (−1) + 2 = 1
x = 0: y = 0 + 2 = 2
x = 1: y = 1 + 2 = 3
x = 2: y = 2 + 2 = 4
x = 3: y = 3 + 2 = 5
x = 4: y = 4 + 2 = 6
x −1 0 1 2 3 4
y 1 2 3 4 5 6
Make a table of values for y = −x + 4. Do the same for the second equation.
x = −1: y = −(−1) + 4 = 1 + 4 = 5
x = 0: y = −0 + 4 = 0 + 4 = 4
x = 1: y = −1 + 4 = −1 + 4 = 3
x = 2: y = −2 + 4 = −2 + 4 = 2
x = 3: y = −3 + 4 = −3 + 4 = 1
x = 4: y = −4 + 4 = −4 + 4 = 0
x −1 0 1 2 3 4
y 5 4 3 2 1 0
Plot both lines on the same grid. Plot each set of points and join them with a ruler. Label each line.
Read the point where the lines cross. This is the point that is on both lines.
The lines cross at (1, 3).

x	−1	0	1	2	3	4
y	1	2	3	4	5	6

x	−1	0	1	2	3	4
y	5	4	3	2	1	0

Check. Check in both equations: y = x + 2 gives 1 + 2 = 3; y = −x + 4 gives −1 + 4 = 3. ✓

Watch out. Use a sharp pencil and a ruler. A thick or wobbly line makes the crossing point hard to read.

Done

FOn the same grid, draw y = 2x − 1 and y = 5 − x for x from 0 to 4. Find the point where the lines cross.

Method. Draw both lines on one grid, then read the coordinates of the point where they cross.

Make a table of values for y = 2x − 1. Substitute each x value. Put negatives in brackets.
x = 0: y = 2 × 0 − 1 = 0 − 1 = −1
x = 1: y = 2 × 1 − 1 = 2 − 1 = 1
x = 2: y = 2 × 2 − 1 = 4 − 1 = 3
x = 3: y = 2 × 3 − 1 = 6 − 1 = 5
x = 4: y = 2 × 4 − 1 = 8 − 1 = 7
x 0 1 2 3 4
y −1 1 3 5 7
Make a table of values for y = −x + 5. Do the same for the second equation.
x = 0: y = −0 + 5 = 0 + 5 = 5
x = 1: y = −1 + 5 = −1 + 5 = 4
x = 2: y = −2 + 5 = −2 + 5 = 3
x = 3: y = −3 + 5 = −3 + 5 = 2
x = 4: y = −4 + 5 = −4 + 5 = 1
x 0 1 2 3 4
y 5 4 3 2 1
Plot both lines on the same grid. Plot each set of points and join them with a ruler. Label each line.
Read the point where the lines cross. This is the point that is on both lines.
The lines cross at (2, 3).

x	0	1	2	3	4
y	−1	1	3	5	7

x	0	1	2	3	4
y	5	4	3	2	1

Check. Check in both equations: y = 2x − 1 gives 2 × 2 − 1 = 3; y = −x + 5 gives −2 + 5 = 3. ✓

Watch out. Use a sharp pencil and a ruler. A thick or wobbly line makes the crossing point hard to read.

Done

FFind the coordinates of the point where the line y = 2x + 1 crosses each of these lines.

(a)the y-axis (x = 0)

(b)the x-axis (y = 0)

(c)the line x = 3

(d)the line y = −3

Done

FThe graph of y = 2x − 1 is drawn. Use the graph to solve each equation.

(a)2x − 1 = 5

(b)2x − 1 = −3

Done

PSThe point P(2, 5) is said to be where the lines y = 3x − 1 and y = x + 3 cross.

(a)Is P on the line y = 3x − 1?

(b)Is P on the line y = x + 3?

(c)Is P the point of intersection of the two lines? Explain.

Done

PSA fence runs along the line x = −2. A straight path follows the line y = 12x + 3. Find the coordinates of the point where the path crosses the fence.

Done

RDo the lines y = 3x + 1 and y = 3x − 4 cross? Give a reason for your answer.

Done

RExplain why two different straight lines cannot cross at two points.

Done

FUse algebra to find the coordinates of the point where each pair of lines cross.

(a)y = 2x + 1 and y = −x + 7

(b)y = 3x − 2 and y = x + 4

(c)y = 4x − 5 and y = −2x + 7

(d)y = 12x + 1 and y = −x + 4

Done

PSThe lines y = x and y = −x + 6 and the x-axis form a triangle. Work out the area of the triangle.

Done

Chapter summary

Skill	Method
Coordinates	(x, y): across first, then up or down. Quadrants 1 to 4 go anticlockwise from the top right.
Midpoint	Add the coordinates and halve: ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2)
Length of a line segment	Subtract to find across and up, then Pythagoras: length² = across² + up²
Horizontal and vertical lines	y = b is horizontal; x = a is vertical
Drawing a line	Table of values, plot, ruler; or use c and the gradient; or the axis intercepts
Gradient	rise ÷ run = (y₂ − y₁) ÷ (x₂ − x₁)
y = mx + c	m is the gradient, c is the y-intercept; rearrange first if needed
Equation from two points	Find m, substitute a point to find c, check with the other point
Parallel lines	Equal gradients
Perpendicular lines (Higher)	Gradients multiply to −1 (negative reciprocal)
Real-life graphs	Gradient = rate of change; y-intercept = starting value; through the origin = direct proportion
Intersections	Read from the graph and check in both equations; Higher: set the right-hand sides equal and solve

Can you…?

I can read and plot coordinates in all four quadrants and find missing corners of shapes.
I can find the midpoint of a line segment, and (Higher) find an end point from the midpoint.
I can find the length of a line segment using Pythagoras’ theorem.
I can write the equations of horizontal and vertical lines.
I can draw a straight-line graph from a table of values.
I can find a gradient from a graph or from two points.
I can use y = mx + c to find the gradient and intercept and to sketch a line.
I can find the equation of a line from a graph, from two points or from a point and a gradient.
I can use gradients to decide if lines are parallel (and, Higher, perpendicular).
I can interpret the gradient and intercept of a real-life graph, and use a graph to find values.
I can find the point where two lines cross.

Key terms

Coordinates: A pair of numbers (x, y) that gives the position of a point: across first, then up or down.
Origin: The point (0, 0) where the axes cross.
Quadrant: One of the four regions the axes divide the grid into.
Midpoint: The point exactly halfway between two points.
Line segment: The part of a straight line between two points.
Gradient: How steep a line is: rise ÷ run.
Rate of change: How much one quantity changes for each 1 unit of another; the gradient of a real-life graph.
y-intercept: The value of y where a line crosses the y-axis (where x = 0).
Linear: Giving a straight-line graph; x appears only to the power 1.
Parallel: Lines that never meet because they have the same gradient.
Perpendicular: Meeting at a right angle (90°); the gradients multiply to −1.
Point of intersection: The point where two lines cross.
Direct proportion: A relationship where doubling one quantity doubles the other; its graph is a straight line through the origin.
Bisect: To cut exactly in half.
Hypotenuse: The longest side of a right-angled triangle, opposite the right angle.

Exercise 12Z Check-up (non-calculator)

Non-calculator

Which quadrant is the point (−3, 5) in?

Done

Find the midpoint of A(−2, 3) and B(6, 9).

Done

Work out the gradient of the line through (1, 2) and (4, 11).

Done

Complete a table of values for y = 2x − 3 for x = −1, 0 and 2.

Key words. A table of values lists some x values and the matching y values. To substitute means to replace a letter with a number.

Method. Put each value of x into the equation and work out y. Do the multiplication before the addition.

Write the rule. The equation tells us how to get y from x: multiply by the gradient, then add the constant.
y = 2x − 3
Substitute each x value in turn. Replace the letter x with the number. Put negative numbers in brackets so you do not lose the sign.
x = −1: y = 2 × (−1) − 3 = −2 − 3 = −5
x = 0: y = 2 × 0 − 3 = 0 − 3 = −3
x = 2: y = 2 × 2 − 3 = 4 − 3 = 1
Write the answers in the table. Each pair (x, y) is a point on the line.
x −1 0 2
y −5 −3 1

x	−1	0	2
y	−5	−3	1

Check. Substitute one value back in a different order to check. ✓

Watch out. Careful with negatives: 2 × (−3) = −6, and −6 + 1 = −5. A negative times a positive is negative.

Done

Write down the gradient and y-intercept of the line y = 5 − 2x.

Done

Find the equation of the line with gradient −2 that passes through (1, 4).

Done

Find the equation of the line parallel to y = 3x + 1 that passes through (0, −2).

Done

Is the point (2, 7) on the line y = 3x + 1?

Done

Work out the length of the line from (0, 0) to (5, 12).

Done

Find the equation of the line through (0, 1) and (2, 5).

Done

Total 54 marks. Allow about 54 minutes.

Exercise 12X Exam-style questions

Non-calculator

FABCD is a parallelogram. A is (−2, 1), B is (1, −2) and C is (5, −2). Write down the coordinates of D.[2 marks]

Done

FA is the point (−3, 4) and B is the point (5, −2). Find the coordinates of the midpoint of AB.[2 marks]

Done

FA is the point (−2, 1) and B is the point (4, 5). Work out the length of AB. Give your answer correct to 1 decimal place.[3 marks]

Done

F(a) Complete a table of values for y = 2x − 3 for values of x from −1 to 4. (b) Draw the graph of y = 2x − 3.[3 marks]

▶ Watch this solution as a video

Method. Draw a table of values, plot the points, then join them with a straight line.

Make a table of values. Two points are enough to draw a straight line, but three or more let you spot a mistake: all the points must line up.
Work out y for each x. Substitute each x into the equation.
x = −1: y = 2 × (−1) − 3 = −2 − 3 = −5
x = 0: y = 2 × 0 − 3 = 0 − 3 = −3
x = 1: y = 2 × 1 − 3 = 2 − 3 = −1
x = 2: y = 2 × 2 − 3 = 4 − 3 = 1
x = 3: y = 2 × 3 − 3 = 6 − 3 = 3
x = 4: y = 2 × 4 − 3 = 8 − 3 = 5
x −1 0 1 2 3 4
y −5 −3 −1 1 3 5
Plot the points. Each pair is a coordinate: across first, then up or down.
(−1, −5) (0, −3) (1, −1) (2, 1) (3, 3) (4, 5)
Join them with a ruler and extend the line. The graph of a linear equation is a straight line, so the points must line up. If one is out of line, check its working.

x	−1	0	1	2	3	4
y	−5	−3	−1	1	3	5

Check. All the plotted points lie on one straight line. A point taken from the line, such as one with a different x, also satisfies the equation. ✓

Watch out. Use a ruler and extend the line across the whole grid. If the points do not line up, one y value is wrong.

Done

FFind the equation of the straight line shown on the grid.[3 marks]

Done

FA straight line passes through the points (1, 5) and (4, 11). Find the equation of the line.[3 marks]

Done

FThe line L has equation y = 3x − 2. Find the equation of the line that is parallel to L and passes through the point (1, 7).[3 marks]

Done

FThe graph shows the cost, £C, of a taxi journey of n miles. (a) Work out the cost of a journey of 4 miles. (b) Rob pays £15. How many miles was his journey? (c) Work out the cost per mile.[4 marks]

(a)the cost of 4 miles

(b)the distance for £15

(c)the cost per mile

Done

FShow that the line y = 2x + 1 is parallel to the line 2y − 4x = 7.[2 marks]

Done

FOn the same grid, draw the graphs of y = x + 1 and y = 5 − x for x from 0 to 5. Write down the coordinates of the point where the two lines cross.[4 marks]

▶ Watch this solution as a video

Method. Draw both lines on one grid, then read the coordinates of the point where they cross.

Make a table of values for y = x + 1. Substitute each x value. Put negatives in brackets.
x = 0: y = 0 + 1 = 1
x = 1: y = 1 + 1 = 2
x = 2: y = 2 + 1 = 3
x = 3: y = 3 + 1 = 4
x = 4: y = 4 + 1 = 5
x = 5: y = 5 + 1 = 6
x 0 1 2 3 4 5
y 1 2 3 4 5 6
Make a table of values for y = −x + 5. Do the same for the second equation.
x = 0: y = −0 + 5 = 0 + 5 = 5
x = 1: y = −1 + 5 = −1 + 5 = 4
x = 2: y = −2 + 5 = −2 + 5 = 3
x = 3: y = −3 + 5 = −3 + 5 = 2
x = 4: y = −4 + 5 = −4 + 5 = 1
x = 5: y = −5 + 5 = −5 + 5 = 0
x 0 1 2 3 4 5
y 5 4 3 2 1 0
Plot both lines on the same grid. Plot each set of points and join them with a ruler. Label each line.
Read the point where the lines cross. This is the point that is on both lines.
The lines cross at (2, 3).

x	0	1	2	3	4	5
y	1	2	3	4	5	6

x	0	1	2	3	4	5
y	5	4	3	2	1	0

Check. Check in both equations: y = x + 1 gives 2 + 1 = 3; y = −x + 5 gives −2 + 5 = 3. ✓

Watch out. Use a sharp pencil and a ruler. A thick or wobbly line makes the crossing point hard to read.

Done

FThe cost of Plan A is £10 plus £2 per GB, and the cost of Plan B is £4 plus £3 per GB. Use the graph to find the amount of data for which the plans cost the same, and say which plan is cheaper for 8 GB.[3 marks]

Done

FA is (0, 1) and B is (4, 3). Find an equation of the line through (2, 5) that is perpendicular to AB.[4 marks]

Done

FA is (0, 1) and B is (4, 3). Find the equation of the perpendicular bisector of AB.[5 marks]

Done

FThe vertices of triangle ABC are A(1, 1), B(5, 3) and C(3, 7). Show that the triangle is isosceles.[4 marks]

Done

FFind the coordinates of the point where the lines y = 3x − 2 and y = x + 4 cross.[4 marks]

Done

FA circle has centre C(2, 1). The point P(6, 4) is on the circle. Find the equation of the tangent to the circle at P.[5 marks]

Done

← Chapter 11: Sequences Chapter 13: Quadratic and other graphs →

GCSE Maths Straight-Line Graphs and Coordinates (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

Coordinates in four quadrants

Coordinates and shapes

Key ideas

Watch out

Exercise 12A Coordinates and shapes on a grid

Midpoints and lengths of line segments

The midpoint

The length of a line segment

Working backwards (Higher)

Key ideas

Watch out

Exercise 12B Midpoints

Exercise 12C Lengths of line segments

Horizontal and vertical lines, and drawing graphs from tables

Drawing a line from its equation

Key ideas

Watch out

Exercise 12D Horizontal and vertical lines, tables of values and drawing lines

Gradient

Finding the gradient from a graph

Finding the gradient from two points

Special cases

Key ideas

Watch out

Exercise 12E Gradient

The equation y = mx + c

Sketching a line using m and c

Lines that are not written as y = …

Finding the axis crossing points

Key ideas

Watch out

Exercise 12F Gradient, intercept and sketching lines

Finding the equation of a straight line

1. From a graph

2. From the gradient and one point

3. From two points

Key ideas

Watch out

Exercise 12G Equations of lines from graphs

Exercise 12H Equations of lines from points and gradients

Parallel and perpendicular lines

Finding a line parallel to another

Perpendicular lines (Higher)

Equation of a perpendicular line (Higher)

Key ideas

Watch out

Exercise 12I Parallel lines

Exercise 12J Perpendicular lines

Real-life graphs and rates of change

What the gradient and intercept mean

Reading values from a graph

Comparing two graphs

Journey graphs

Direct and inverse proportion graphs

Key ideas

Watch out

Exercise 12K Conversion graphs, rates and proportion

Exercise 12L Fares, tariffs and comparing lines

Where lines cross

Solving equations with a graph

Key ideas

Watch out

Exercise 12M Intersections of lines

Chapter summary

Can you…?

Key terms

Exercise 12Z Check-up (non-calculator)

Exercise 12X Exam-style questions