## Gradients of Straight Line Graphs

Equations featuring x and y (and no x^3 or y^2 etc.) are **straight lines**.

The **gradient** of a line is a measure of how **steep** it is. If the gradient is small, the slope of the line will be very gradual, but if the gradient is big, the line will be quite steep. You are required to know how to **calculate the gradient** from two possible circumstances:

- You are given the line drawn on a graph.
- You are given two coordinates and told that a line passes through both of them.

## Check out the MME Learning Portal

Online exams, practice questions and revision videos for every GCSE level 9-1 topic!

**No fees, no trial period, just totally free access to the UK's best GCSE maths revision platform.**

**Finding the Gradient of a Straight Line **

**Example:** Find the gradient of the line shown:

In general, if we have two coordinates (\textcolor{blue}{x_1}, \textcolor{red} {y_1}) and (\textcolor{blue}{x_2}, \textcolor{red}{y_2}) then the **gradient of the line** that passes through them is,

\begin{aligned} \text{Gradient } &= \dfrac{\text{change in }\textcolor{red}{y}}{\text{change in }\textcolor{blue}{x}} = \dfrac{\textcolor{red}{y_2 - y_1}}{\textcolor{blue}{x_2 - x_1}} \\ \\ & = \dfrac{2 - (-1)}{2 - 0} = \dfrac{3}{2} \end{aligned}

**Positive vs Negative Gradients**

- Positive gradient: \textcolor{red}{y} value
as the \textcolor{blue}{x} value*increases***increases**. - Negative gradient: \textcolor{red}{y} value
as the \textcolor{blue}{x} value*decreases***increases**.

**Horizontal and Vertical Lines**

**Horizontal lines:** y = a (constant y value)

**Vertical lines:** x=a (constant x value)

**Example:** Draw the graphs of x=1 and y=2.

When drawing such lines, find the coordinate on the correct axis, i.e. for x=1, find (1,0) and draw a straight line perpendicular to that axis which results in a vertical line.

Similarly for y=2, find (0,2) and draw a straight line perpendicular to that axis, which in this case is a horizontal line.

## Example**:**** Find the Gradient of a Line Through Two Points**

Work out the **gradient of the straight line** that passes through the points (2, 3) and (-10, 6).

**Step 1:** **pick one of the points** and subtract its x and y coordinates from the other point’s x and y coordinates respectively.

\text{change in } y = (y_1 - y_2) = (6-3) = 3

\text{change in } x = (x_1 - x_2) = (-10-2) = -12

**Step 2:** **Substitute** in to the formula,

\text{gradient } = \dfrac{\text{change in } y}{\text{change in } x} = \dfrac{3}{-12} = -\dfrac{1}{4}

### Take an Online Exam

#### Gradients of Straight Line Graphs Online Exam

### Example Questions

**Question 1:** Calculate the gradient of the line drawn below.

**[2 marks]**

To do this, you want to pick 2 points on the graph that the line passes through. It’s best, if you can, to pick two points where the coordinates are easy to read off. Here, we picked (2, 1) and (4, 5), as seen on the graph.

Once you’ve done this, draw the right-angled triangle as pictured with dotted lines. Then, the change in x is the width of the base of that triangle, whilst the change in y is the height.

Therefore, we get

\text{gradient } = \dfrac{\text{change in }y}{\text{change in }x}= \dfrac{4}{2} = 2

**Question 2:** Calculate the gradient of the line drawn below.

**[2 marks]**

We must find two points that the line passes through and draw a right-angled triangle underneath, so we can identify the change in x to be the base and the change in y to be the height. This looks like

Now, given that this is a downwards slope, it must have a negative gradient. So, we get

\text{gradient } = -\dfrac{3}{1} = -3

Note: you could’ve used a different triangle at different points on the line – this is fine, as long as you got the correct answer of - 3.

**Question 3:** Find the gradient of the line that passes through (-8, -1) and (2, -6).

**[2 marks]**

To find the gradient, we’ll subtract the values of second coordinate from those of the first, and divide the difference in the y values by the difference in the x values:

\text{gradient } = \dfrac{-1 - (-6)}{-8 - 2} = \dfrac{5}{-10} = -\dfrac{1}{2}

**Question 4:** Draw the lines y=-2 and x=-3 on a single graph.

**[2 marks]**

When drawing lines of the form x=a or y=b, find the coordinate on the correct axis and draw a straight line perpendicular to that axis.

For y=-2, find (0,-2) and draw a straight line perpendicular to that axis which is a horizontal line.

For x=-3, find (-3,0) and draw a straight line perpendicular to that axis which is a vertical line.

### Worksheets and Exam Questions

#### (NEW) Gradients of Straight Line Graphs Exam Style Questions - MME

Level 1-3 New Official MME### Drill Questions

#### Straight Line and Gradients - Drill Questions

#### Gradients of Straight Line Graphs - Drill Questions

#### Straight Line Graphs - Drill Questions

#### Straight Line Graphs 2 - Drill Questions

#### Graph Exam Questions (harder)

### Learning resources you may be interested in

We have a range of learning resources to compliment our website content perfectly. Check them out below.