Gradient of a Linear Graph

Measure how steep a straight line is: draw a right-angled step on the line, count its vertical change and its horizontal change in the units of the axes, and divide.

Before you start

A point is fixed by its coordinates $(x, y)$ : the $x$ -coordinate is read on the horizontal axis and the $y$ -coordinate on the vertical axis (Lesson 10.1). On some graphs one square stands for 2, 5 or 10 units, so read the numbers on the axis, not the squares.
The graph of a linear function $y = ax + b$ is a straight line, drawn through the points of a table of values (Lesson 10.4).
A fraction is written in its simplest form, and a negative number divided by a positive number is negative: $\frac{6}{4} = \frac{3}{2}$ and $\frac{-8}{4} = -2$ .
A rate compares two quantities measured in different units, such as $4 per bowl or $60$ km/h.

What Gradient Measures

Some straight lines rise gently, some rise steeply and some fall. The gradient of a line is a number that measures how steep it is: how many units the line goes up (or down) for every $1$ unit it goes across to the right.

To find it, choose two points on the line where it passes exactly through a corner of the grid. From the left-hand point draw a right-angled step: go across to the right until you are directly below (or above) the other point, then go up (or down) to it. The distance across is the horizontal change; the distance up or down is the vertical change.

Gradient of a straight line

$\text{gradient} = \frac{\text{vertical change}}{\text{horizontal change}}$

Both changes are counted in the units of the axes, going from the left-hand point to the right-hand point. So the horizontal change is always positive. The vertical change is positive when the step goes up and negative when it goes down.

Example 1: Finding a gradient with a step

Points $A(1, 2)$ and $B(4, 8)$ lie on a straight line. Draw the line and find its gradient.

Solution

Plot $A$ and $B$ , draw the line through them, and draw the step from $A$ , the left-hand point.

Across: from $A(1, 2)$ go right to $(4, 2)$ , directly below $B$ . The $x$ -coordinate goes from $1$ to $4$ , so the horizontal change is $4 - 1 = 3$ .

Up: from $(4, 2)$ go up to $B(4, 8)$ . The $y$ -coordinate goes from $2$ to $8$ , so the vertical change is $8 - 2 = 6$ .

$\text{gradient} = \frac{6}{3} = 2$

The gradient of the line is $2$ : it goes up $2$ units for every $1$ unit across.

Positive, Negative and Zero Gradients

Read a line from left to right, as you read a sentence. Whether it goes up, goes down or stays level decides the sign of its gradient.

The sign of a gradient

A line that goes up from left to right has a positive gradient.

A line that goes down from left to right has a negative gradient: its step goes down, so its vertical change is negative.

A horizontal line has gradient $0$ : its vertical change is $0$ .

Example 2: A line going down

Points $C(1, 6)$ and $D(3, 2)$ lie on a straight line. Find the gradient of the line.

Solution

Draw the line and the step from $C$ , the left-hand point. The line goes down, so the step goes across and then down.

Across: from $C(1, 6)$ go right to $(3, 6)$ , directly above $D$ . The horizontal change is $3 - 1 = 2$ .

Down: from $(3, 6)$ go down to $D(3, 2)$ . The $y$ -coordinate falls from $6$ to $2$ , so the vertical change is $2 - 6 = -4$ .

$\text{gradient} = \frac{-4}{2} = -2$

The gradient is $-2$ : the line goes down $2$ units for every $1$ unit across.

Example 3: A horizontal line

Points $P(1, 3)$ and $Q(5, 3)$ lie on the line $y = 3$ . Find its gradient.

Solution

Every point on the line has $y$ -coordinate $3$ , so the line is horizontal.

Across: from $P$ to $Q$ the horizontal change is $5 - 1 = 4$ .

Up or down: the $y$ -coordinate stays at $3$ , so there is no step up or down: the vertical change is $3 - 3 = 0$ .

$\text{gradient} = \frac{0}{4} = 0$

The line $y = 3$ has gradient $0$ , as every horizontal line does.

Any Two Points Give the Same Gradient

A straight line is equally steep all along its length. So it does not matter which two points on it you choose: a small step and a big step on the same line give the same gradient. Choose points that make the counting easy.

Example 4: A small step and a big step

The line $y = \frac{1}{2}x + 1$ passes through $A(2, 2)$ , $B(4, 3)$ and $C(8, 5)$ . Find its gradient using $A$ and $B$ , then using $B$ and $C$ , then using $A$ and $C$ .

Solution

Small step, $A$ to $B$ : across $4 - 2 = 2$ , up $3 - 2 = 1$ . Gradient $= \frac{1}{2}$ .

Big step, $B$ to $C$ : across $8 - 4 = 4$ , up $5 - 3 = 2$ . Gradient $= \frac{2}{4} = \frac{1}{2}$ .

From $A$ to $C$ (not drawn): across $8 - 2 = 6$ , up $5 - 2 = 3$ . Gradient $= \frac{3}{6} = \frac{1}{2}$ .

All three steps give the gradient $\frac{1}{2}$ : the line goes up $1$ unit for every $2$ units across.

Axes with Different Scales

On many graphs one square across and one square up do not stand for the same amount. Then counting squares gives the wrong gradient: the changes must be read from the numbers on the axes.

Count in units, not in squares

Before you count, check what one square stands for on each axis. A step of $3$ squares up is $3$ units only when one square up is $1$ unit.

Example 5: Unequal scales

The graph shows the line $y = 4x + 2$ , with the points $A(1, 6)$ and $B(4, 18)$ on it. On the $x$ -axis one square is $1$ unit; on the $y$ -axis one square is $2$ units. Find the gradient of the line.

Solution

Across: from $A(1, 6)$ to $(4, 6)$ . That is $3$ squares, and each square across is $1$ unit, so the horizontal change is $4 - 1 = 3$ .

Up: from $(4, 6)$ to $B(4, 18)$ . That is $6$ squares, but each square up is $2$ units, so the vertical change is $18 - 6 = 12$ .

$\text{gradient} = \frac{12}{3} = 4$

Counting squares would give $\frac{6}{3} = 2$ , which is wrong: the line goes up $12$ units, not $6$ .

The gradient of the line is $4$ .

Comparing Steepness

The sign of a gradient tells you which way a line goes. Its size, ignoring the sign, tells you how steep it is.

Steepness

On the same axes, with the same scale on both, the larger the size of the gradient (ignoring its sign), the steeper the line.

Gradients $3$ and $-3$ are equally steep: one line goes up $3$ units for every $1$ across and the other goes down $3$ units.

So a larger gradient does not always mean a steeper line: $-3$ is smaller than $1$ , but a line with gradient $-3$ is steeper than a line with gradient $1$ .

On axes with different scales a line can look gentler or steeper than it is (the line of Example 5 looks as steep as a line of gradient $2$ ), so compare the gradients, not the pictures.

Example 6: Which line is steepest?

The lines $y = \frac{1}{2}x$ , $y = x$ , $y = 3x$ and $y = -3x$ are drawn below from the origin to the right.

(a) Find the gradient of each line.

(b) Which two lines are equally steep?

Solution

(a) From the origin, follow each line to a corner of the grid and use that step.

$y = \frac{1}{2}x$ : from $(0, 0)$ to $(2, 1)$ , across $2$ and up $1$ . Gradient $= \frac{1}{2}$ .

$y = x$ : from $(0, 0)$ to $(1, 1)$ , across $1$ and up $1$ . Gradient $= \frac{1}{1} = 1$ .

$y = 3x$ : from $(0, 0)$ to $(1, 3)$ , across $1$ and up $3$ . Gradient $= \frac{3}{1} = 3$ .

$y = -3x$ : from $(0, 0)$ to $(1, -3)$ , across $1$ and down $3$ . Gradient $= \frac{-3}{1} = -3$ .

(b) The gradients $3$ and $-3$ have the same size, $3$ . The line $y = 3x$ goes up $3$ units for each unit across and $y = -3x$ goes down $3$ units, so they are equally steep.

(a) $\frac{1}{2}$ , $1$ , $3$ and $-3$ . (b) $y = 3x$ and $y = -3x$ . (c) $y = 3x$ and $y = -3x$ (equally steep), then $y = x$ , then $y = \frac{1}{2}x$ .

Gradient from Two Points or from a Table

When you are given two points but no grid, make a quick sketch: mark the two points roughly where they belong, draw the step from the left-hand point, and work out each change by subtracting coordinates.

The two changes from coordinates

Horizontal change $=$ the $x$ -coordinate of the right-hand point $-$ the $x$ -coordinate of the left-hand point.

Vertical change $=$ the $y$ -coordinate of the right-hand point $-$ the $y$ -coordinate of the left-hand point.

Example 7: Two given points

Find the gradient of the line through $P(-1, 7)$ and $Q(3, -1)$ .

Solution

Sketch the line with its step. $P$ is the left-hand point, since $-1$ is less than $3$ .

Horizontal change: from $x = -1$ to $x = 3$ , so $3 - (-1) = 3 + 1 = 4$ .

Vertical change: from $y = 7$ down to $y = -1$ , so $-1 - 7 = -8$ . It is negative because the step goes down.

$\text{gradient} = \frac{-8}{4} = -2$

The gradient of the line through $P$ and $Q$ is $-2$ .

In a table of values for a straight line, each column is a point on the line. So any two columns give a step: the change in $x$ is the horizontal change and the change in $y$ is the vertical change.

Example 8: Gradient from a table of values

The table shows values of $x$ and $y$ for a straight line. Find the gradient of the line.

$x$	$-2$	$0$	$2$	$4$
$y$	$-1$	$2$	$5$	$8$

Solution

From each column to the next, $x$ increases by $2$ and $y$ increases by $3$ : a step $2$ across and $3$ up, as the graph of the table shows.

$\text{gradient} = \frac{3}{2}$

Check with the first and last columns: $x$ goes from $-2$ to $4$ , a change of $6$ ; $y$ goes from $-1$ to $8$ , a change of $9$ . Gradient $= \frac{9}{6} = \frac{3}{2}$ .

The gradient of the line is $\frac{3}{2}$ .

Drawing a Line with a Given Gradient

To draw a line through a given point with a given gradient, write the gradient as $\frac{\text{vertical change}}{\text{horizontal change}}$ and use it as a step: from the point go across to the right, then up (or down), and mark a new point. Take the same step once more as a check, then join the points with a ruler.

A whole-number gradient is a step of $1$ across: a gradient of $-2$ is $\frac{-2}{1}$ , so $1$ across and $2$ down.

Example 9: Drawing a line with a negative gradient

Draw the line through $A(1, 5)$ with gradient $-\frac{3}{2}$ .

Solution

Write the gradient as $\frac{-3}{2}$ : for every $2$ units across to the right, the line goes $3$ units down.

From $A(1, 5)$ : across $2$ to $x = 3$ , then down $3$ to $y = 2$ . Mark $B(3, 2)$ .

From $B(3, 2)$ : across $2$ to $x = 5$ , then down $3$ to $y = -1$ . Mark $C(5, -1)$ .

Join $A$ , $B$ and $C$ with a ruler and extend the line.

Check: from $A$ to $C$ the line goes across $4$ and down $6$ , and $\frac{-6}{4} = -\frac{3}{2}$ .

The line passes through $A(1, 5)$ , $B(3, 2)$ and $C(5, -1)$ , going down from left to right.

Gradient as a Rate

When the axes stand for real quantities, the gradient tells you how fast one quantity changes as the other increases: it is a rate. The scales of such graphs are often different, so count in the units of each axis.

Gradient as a rate

The unit of the gradient is the unit of the vertical axis per unit of the horizontal axis: dollars per bowl, litres per minute, kilometres per hour.

Example 10: Cost per item

A hawker stall sells fishball noodles at the same price for every bowl. The graph shows the cost of $1$ to $6$ bowls. One square across is $1$ bowl; one square up is $2.

(a) Why are the points not joined?

(b) Find the gradient and say what it stands for.

Solution

(a) A customer buys a whole number of bowls: there is no cost for $2\frac{1}{2}$ bowls. So the points are plotted but not joined.

(b) The points lie on a straight line, so a step between any two of them gives its gradient. Use the points for $1$ bowl and $5$ bowls.

Across: from $1$ bowl to $5$ bowls, $5 - 1 = 4$ bowls.

Up: from $4 to $20, $20 - 4 = 16$ dollars ( $8$ squares of $2 each).

$\text{gradient} = \frac{16}{4} = 4$

(a) Only whole numbers of bowls are sold. (b) The gradient is $4$ dollars per bowl: each bowl of fishball noodles costs $4.

Example 11: Litres per minute

Water flows into a tank at a steady rate. The graph shows the volume of water, $V$ litres, in the tank $t$ minutes after the tap is turned on. One square across is $1$ minute; one square up is $5$ litres.

(a) Find the gradient of the line, with its unit, and say what it stands for.

(b) How much water was in the tank when the tap was turned on?

Solution

The graph is an unbroken line because time passes continuously and the volume rises smoothly with it.

(a) The line passes through $(2, 30)$ and $(6, 50)$ at corners of the grid. Draw the step between them.

Across: $6 - 2 = 4$ minutes. Up: $50 - 30 = 20$ litres. (The step is $4$ squares across and $4$ squares up, but the squares stand for different amounts.)

$\text{gradient} = \frac{20}{4} = 5$

(b) At $t = 0$ the line is at $V = 20$ .

(a) The gradient is $5$ litres per minute: water flows in at $5$ litres every minute. (b) There were $20$ litres in the tank at the start.

Example 12: A distance–time graph

A coach travels from Singapore to Malacca at a constant speed. The graph shows the distance, $d$ km, it has travelled $t$ hours after it set off. One square across is half an hour; one square up is $20$ km. Find the gradient of the graph and say what it stands for.

Solution

The line passes through $(1, 60)$ and $(3, 180)$ at corners of the grid. Draw the step between them.

Across: $3 - 1 = 2$ hours. Up: $180 - 60 = 120$ km.

$\text{gradient} = \frac{120}{2} = 60$

The unit is kilometres per hour: the coach covers $60$ km in every hour.

The gradient is $60$ km/h: it is the speed of the coach.

The Gradient of $y = ax + b$

Look back at some of the lines in this lesson, their equations and the gradients you found:

Line	Equation	Gradient
Example 1	$y = 2x$	$2$
Example 2	$y = -2x + 8$	$-2$
Example 3	$y = 3$ , that is $y = 0x + 3$	$0$
Example 4	$y = \frac{1}{2}x + 1$	$\frac{1}{2}$
Example 5	$y = 4x + 2$	$4$
Example 7	$y = -2x + 5$	$-2$

The equations of Examples 1, 2 and 7 were not given there. Check each by substituting its two points: $2 = 2(1)$ and $8 = 2(4)$ ; $6 = -2(1) + 8$ and $2 = -2(3) + 8$ ; $7 = -2(-1) + 5$ and $-1 = -2(3) + 5$ .

Each time, the gradient is the number in front of $x$ . This is no accident: in $y = ax + b$ , each time $x$ increases by $1$ , $ax$ increases by $a$ , so $y$ changes by $a$ for every $1$ unit across.

Gradient and equation

The graph of $y = ax + b$ is a straight line with gradient $a$ .

It crosses the $y$ -axis at $(0, b)$ , because $y = b$ when $x = 0$ .

In Example 11 the line is $V = 5t + 20$ : the gradient $5$ is the rate of flow in litres per minute, and $20$ litres is the volume at the start.

Common Mistakes

Counting squares on unequal scales. In Example 5, counting squares gives half the true gradient. Read the changes from the axis numbers.
Turning the fraction upside down. The gradient is the vertical change divided by the horizontal change, not the other way round.
Losing the minus sign. Count across from left to right. If the line goes down, the vertical change is negative and so is the gradient.
"The bigger the gradient, the steeper the line." This is false for negative gradients: $-4$ is smaller than $2$ , but a line with gradient $-4$ is steeper. Compare the sizes, ignoring the signs.
Using points that are not exactly on the line. Choose points where the line passes through a corner of the grid, or points given in the question or the table.
Leaving out the unit of a rate. A gradient of $5$ on a volume–time graph is $5$ litres per minute.

Practice Questions

Try each question before you open its solution. For every gradient, draw or sketch the step.

Question 1

Find the gradient of the line shown. The points $A$ and $B$ lie on the line at corners of the grid.

ShowHide solution

Read the points: $A(1, 1)$ and $B(3, 5)$ . Draw the step from $A$ .

Horizontal change $= 3 - 1 = 2$ . Vertical change $= 5 - 1 = 4$ (up).

$\text{gradient} = \frac{4}{2} = 2$

The gradient of the line is $2$ .

Question 2

Find the gradient of the line shown, which passes through the points $A$ and $B$ .

ShowHide solution

Read the points: $A(-4, 5)$ is the left-hand point and $B(2, 2)$ the right-hand point.

Horizontal change $= 2 - (-4) = 6$ . The line goes down, so the vertical change is $2 - 5 = -3$ .

$\text{gradient} = \frac{-3}{6} = -\frac{1}{2}$

The gradient of the line is $-\frac{1}{2}$ .

Question 3

Sketch the line through each pair of points with its step, and find its gradient.

(a) $(2, 5)$ and $(6, 13)$

(b) $(-2, 9)$ and $(4, 0)$

ShowHide solution

(a) From $(2, 5)$ go across to $(6, 5)$ , then up to $(6, 13)$ . Horizontal change $= 6 - 2 = 4$ ; vertical change $= 13 - 5 = 8$ . Gradient $= \frac{8}{4} = 2$ .

(b) The left-hand point is $(-2, 9)$ . Go across to $(4, 9)$ , then down to $(4, 0)$ .

Horizontal change $= 4 - (-2) = 6$ ; vertical change $= 0 - 9 = -9$ . Gradient $= \frac{-9}{6} = -\frac{3}{2}$ .

(c) Both points have $y$ -coordinate $-2$ , so the line is horizontal. Horizontal change $= 5 - (-3) = 8$ ; vertical change $= -2 - (-2) = 0$ . Gradient $= \frac{0}{8} = 0$ .

(a) $2$ (b) $-\frac{3}{2}$ (c) $0$

Question 4

The table gives points on a straight line. Find the gradient of the line.

$x$	$1$	$3$	$5$	$7$
$y$	$7$	$6$	$5$	$4$

ShowHide solution

From each column to the next, $x$ increases by $2$ and $y$ decreases by $1$ : a step $2$ across and $1$ down.

$\text{gradient} = \frac{-1}{2} = -\frac{1}{2}$

Check with the first and last columns: across $7 - 1 = 6$ , vertical change $4 - 7 = -3$ , and $\frac{-3}{6} = -\frac{1}{2}$ .

The gradient of the line is $-\frac{1}{2}$ .

Question 5

Five lines have these gradients. Line $P$ : $2$ . Line $Q$ : $-4$ . Line $R$ : $\frac{1}{2}$ . Line $S$ : $-2$ . Line $T$ : $0$ .

(a) Which lines go down from left to right? Which line is horizontal?

(b) Which two lines are equally steep?

ShowHide solution

(a) A negative gradient means the line goes down from left to right: $Q$ and $S$ . A gradient of $0$ means a horizontal line: $T$ .

(b) Ignoring the signs, the sizes of the gradients are $2$ , $4$ , $\frac{1}{2}$ , $2$ and $0$ . Lines $P$ and $S$ both have size $2$ : $P$ goes up $2$ units for every $1$ across and $S$ goes down $2$ units, so they are equally steep.

(c) Order the sizes from the largest to the smallest: $4$ ( $Q$ ), $2$ ( $P$ and $S$ ), $\frac{1}{2}$ ( $R$ ), $0$ ( $T$ ).

(a) $Q$ and $S$ go down; $T$ is horizontal. (b) $P$ and $S$ . (c) $Q$ ; then $P$ and $S$ , equally steep; then $R$ ; then $T$ .

Question 6

The graph shows a straight line through the points $A$ and $B$ . One square across is $1$ unit; one square up is $5$ units. Wei Ming counts $2$ squares across and $2$ squares up from $A$ to $B$ , and says that the gradient is $1$ .

(a) Explain his mistake.

(b) Find the correct gradient.

ShowHide solution

(a) The squares do not stand for the same amount on the two axes: one square up is $5$ units. He counted squares instead of units.

(b) Read the points: $A(1, 10)$ and $B(3, 20)$ .

Horizontal change $= 3 - 1 = 2$ . Vertical change $= 20 - 10 = 10$ : $2$ squares of $5$ units each.

$\text{gradient} = \frac{10}{2} = 5$

(a) He counted squares, but each square up is $5$ units. (b) The gradient is $5$ .

Question 7

(a) Draw the line through $(0, 2)$ with gradient $3$ .

(b) Draw the line through $(1, 1)$ with gradient $\frac{2}{3}$ .

For each line, give the coordinates of two more points on it.

ShowHide solution

(a) A gradient of $3$ is $\frac{3}{1}$ : $1$ across and $3$ up. From $(0, 2)$ go to $(1, 5)$ , then to $(2, 8)$ . Join the points.

(b) A gradient of $\frac{2}{3}$ means $3$ across and $2$ up. From $(1, 1)$ go to $(4, 3)$ , then to $(7, 5)$ . Join the points.

(a) For example $(1, 5)$ and $(2, 8)$ . (b) For example $(4, 3)$ and $(7, 5)$ .

Question 8

A money changer at Raffles Place uses the graph below to change Singapore dollars (S$) into Malaysian ringgit (RM) at a fixed rate. One square across is S$5; one square up is RM10.

(a) Use the points $(25, 80)$ and $(50, 160)$ on the line to find its gradient.

(b) What does the gradient stand for?

ShowHide solution

(a) Draw the step from $(25, 80)$ to $(50, 160)$ .

Across: $50 - 25 = 25$ Singapore dollars. Up: $160 - 80 = 80$ ringgit.

$\text{gradient} = \frac{80}{25} = 3.2$

(b) The gradient is $3.2$ ringgit per Singapore dollar: the money changer gives RM3.20 for every S$1.

(a) $3.2$ ringgit per Singapore dollar (b) The rate of exchange: S$1 = RM3.20. (c) RM128

Question 9

Siti plays a game on her phone, and its battery runs down steadily. The graph shows the charge left in the battery, as a percentage of a full charge, $t$ hours after she starts. One square across is $1$ hour; one square up is 10%.

(a) What was the charge at the start?

(b) Find the gradient of the line, with its unit.

(d) After how many hours is the battery flat?

ShowHide solution

(a) At $t = 0$ the line is at 90%.

(b) Use the points $(2, 60)$ and $(4, 30)$ , at corners of the grid.

Across: $4 - 2 = 2$ hours. The line goes down, so the vertical change is $30 - 60 = -30$ , in percent of a full charge.

$\text{gradient} = \frac{-30}{2} = -15$

The unit is the unit up per unit across: percent per hour. So the gradient is $-15$ % per hour.

(d) The line meets the time axis at $t = 6$ : the battery is flat after 6 hours. Check: 90% at 15% an hour lasts $90 \div 15 = 6$ hours.

(a) 90% (b) $-15$ % per hour (c) The charge falls by 15% of a full charge every hour. (d) After $6$ hours

Question 10

Priya uses her EZ-Link card for bus rides that all cost the same fare. The table shows the balance on her card after $n$ rides.

Number of rides, $n$	$0$	$2$	$4$	$6$
Balance ($)	$20.00$	$17.40$	$14.80$	$12.20$

(a) Show that the points lie on a straight line.

(b) Find the gradient of the graph of the balance against $n$ , and say what it means.

ShowHide solution

(a) For every $2$ more rides the balance goes down by the same amount: $20.00 - 17.40 = 2.60$ , $17.40 - 14.80 = 2.60$ and $14.80 - 12.20 = 2.60$ . Equal steps in $n$ give equal steps in the balance, so the points lie on a straight line.

(b) From one column to the next the step is $2$ rides across and $2.60 down, so the vertical change is $-2.60$ .

$\text{gradient} = \frac{-2.60}{2} = -1.30$

The balance falls by $1.30 for each ride: each bus ride costs $1.30.

(c) After $11$ rides the balance is $20.00 - 11 \times 1.30 = 20.00 - 14.30 = 5.70$ dollars. After $12$ rides it is $5.70 - 1.30 = 4.40$ dollars, which is below $5.

(a) The balance falls by $2.60 for every 2 rides, so the steps are equal and the points lie on a straight line. (b) $-1.30$ dollars per ride: each ride costs $1.30. (c) After $12$ rides (balance $4.40).

Question 11

Mdm Lim's class orders printed T-shirts for Sports Day. The cost, $C$ dollars, of $x$ T-shirts is $C = 8x + 150$ .

(a) Copy and complete the table.

$x$	$0$	$10$	$20$	$30$
$C$

(b) Use the table to find the gradient of the graph of $C$ against $x$ .

ShowHide solution

(a) $C = 8(0) + 150 = 150$ ; $C = 8(10) + 150 = 230$ ; $C = 8(20) + 150 = 310$ ; $C = 8(30) + 150 = 390$ .

$x$	$0$	$10$	$20$	$30$
$C$	$150$	$230$	$310$	$390$

(b) For every $10$ more T-shirts, $C$ goes up by $80$ : a step $10$ across and $80$ up.

$\text{gradient} = \frac{80}{10} = 8$

(c) The gradient $8$ is the number in front of $x$ , as for every line $y = ax + b$ . It is the rate: $8 for each T-shirt. The $150$ is the value of $C$ when $x = 0$ : a fixed charge of $150, paid however many T-shirts are ordered.

(a) $150$ , $230$ , $310$ , $390$ (b) $8$ (c) $8 per T-shirt, and a fixed charge of $150.

Question 12

A straight line passes through $A(-2, 1)$ and $B(4, k)$ . Its gradient is $\frac{3}{2}$ .

(a) Find $k$ .

(b) Show that the point $C(2, 7)$ lies on the line.

ShowHide solution

(a) Sketch the step from $A$ , the left-hand point. Horizontal change $= 4 - (-2) = 6$ .

A gradient of $\frac{3}{2}$ means $3$ up for every $2$ across. Across $6$ is $3$ lots of $2$ , so the line goes up $3 \times 3 = 9$ .

So $k = 1 + 9 = 10$ .

(b) From $A(-2, 1)$ to $C(2, 7)$ : across $2 - (-2) = 4$ and up $7 - 1 = 6$ , so the gradient from $A$ to $C$ is $\frac{6}{4} = \frac{3}{2}$ .

This is the gradient of the line. Only one straight line passes through $A$ with gradient $\frac{3}{2}$ , so $C$ lies on it.

(a) $k = 10$ (b) The step from $A$ to $C$ has gradient $\frac{3}{2}$ , the same as the line, so $C$ is on the line.

Key Points

The gradient of a straight line measures its steepness: $\text{gradient} = \frac{\text{vertical change}}{\text{horizontal change}}$ .
Find it with a right-angled step between two points on the line, going from left to right.
Count both changes in the units of the axes, not in squares, above all when the scales differ.
Up from left to right: positive gradient. Down: negative gradient. Horizontal: gradient $0$ .
Any two points on a straight line give the same gradient.
On the same axes, with the same scale on both, the larger the size of the gradient, ignoring its sign, the steeper the line; $3$ and $-3$ are equally steep.
To draw a line with a given gradient, step across and up (or down) from the given point, twice, and join.
On a graph from a real setting the gradient is a rate, with the unit of the vertical axis per unit of the horizontal axis.
The graph of $y = ax + b$ has gradient $a$ and crosses the $y$ -axis at $(0, b)$ .

Before you start

What Gradient Measures

Gradient of a straight line

Example 1: Finding a gradient with a step

Positive, Negative and Zero Gradients

The sign of a gradient

Example 2: A line going down

Example 3: A horizontal line

Any Two Points Give the Same Gradient

Example 4: A small step and a big step

Axes with Different Scales

Count in units, not in squares

Example 5: Unequal scales

Comparing Steepness

Steepness

Example 6: Which line is steepest?

Gradient from Two Points or from a Table

The two changes from coordinates

Example 7: Two given points

Example 8: Gradient from a table of values

Drawing a Line with a Given Gradient

Example 9: Drawing a line with a negative gradient

Gradient as a Rate

Gradient as a rate

Example 10: Cost per item

Example 11: Litres per minute

Example 12: A distance–time graph

The Gradient of y=ax+by = ax + by=ax+b

Gradient and equation

Common Mistakes

Practice Questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Key Points

The Gradient of $y = ax + b$