## What you need to know

To be able to draw a straight line, you need to know some things about it. When asked to draw a straight line, you may be given either:

- A table of x, y values/a list of a coordinates the line passes through, or
- The equation of the line, in the form y = mx + c.

Make sure you are happy with

## Drawing straight line graphs – Table of Values Method

Draw a graph for the line y = 2x - 3.

**Step 1 –** Construct a table with suitable x values

Step 2 – Find the values of y for each x value.

To work out the missing values, we use the equation like a formula, since the coordinates must satisfy the equation if the line is to pass through them. Substituting the values in the table in, we get the following.

\text{When } x = -2, \text{ we get } y = (2\times-2) - 3 = -7

\text{When } x = -2, \text{ we get } y = (2\times0) - 3 = -3

\text{When } x = -2, \text{ we get } y = (2\times2) - 3 = 1

\text{When } x = -2, \text{ we get } y = (2\times-4) - 3 = 5

So, we know that the line passes through

(-2, -7), (0, -3), (1, -1)\text{, and }(4, 5)

Now all that remains is to plot them on a pair of axes and draw a straight line through them. The result should like the graph to the right.

## Drawing Straight line graphs – \text{Using }y=mx+c

Plot the straight-line graph with equation 3y + 9x = 12.

Firstly, subtract 6x from both sides so that the y term is on its own.

3y = -9x + 12

Then, divide both sides by 3 to get

y = -3x + 4

This is the desired form for a straight-line equation. Now we know that the y-intercept is at (0, 4), and the gradient is -3. Therefore, as the x value increases by 1, the y value __decreases__ (due to negative gradient) by 3. Using this information, we should get a graph that looks a little like (well, exactly like) this:

### Example Questions

**Question 1:** Below is a table of coordinates of the line y = \dfrac{1}{2}x + 5. Complete the table, then plot the points and the straight-line.

To find the missing value, substitute the given values into the equation.

\text{When } x = -1, \text{ we get } y = \dfrac{1}{2} \times (-1) + 5 = 4.5

\text{When } x = 2, \text{ we get } y = \dfrac{1}{2} \times (2) + 5 = 6

\text{When } y = 7, \text{ we get } 7 = \dfrac{1}{2}x + 5

Subtract 5 from both sides of this equation to get

2 = \dfrac{1}{2}x

Multiplying both sides by 2 we immediately get x = 4. The completed table looks like:

Plotting these points and using them to draw the graph should look like:

**Question 2:** Plot the graph of the equation 2y + 1 = 8x

Let’s rearrange this equation. Subtract 1 from both sides:

2y = 8x - 1

Then, divide both sides by 2:

y = 4x - \dfrac{1}{2}

So, the y-intercept is -\frac{1}{2}, and the gradient is 4 – so each time x increases by 1, y increases by 4.

This is enough information to draw the graph. The result should look like the figure below.

**Question 3:** Plot the graph of the equation y - \dfrac{1}{2} = -\dfrac{1}{2}x

Rearranging this equation to be in the form y = mx +c, by adding 0.5 to both sides,

y = -0.5x + 0.5

So, the y-intercept is \dfrac{1}{2}, and the gradient is -\dfrac{1}{2} – so each time x increases by 1, y decreases by 0.5

The result should look,

**Question 4:** Plot the graph of the equation 2y + 2 = 3x

We can rearrange this equation by subtracting 2 from both sides:

2y = 3x - 2

Then, dividing both sides by 2:

y = \dfrac{3}{2}x - 1

So, the y-intercept is -1, and the gradient is \dfrac{3}{2} – so each time x increases by 1, y increases by 1.5

The result should look,

**Question 5:** Plot the graph of the equation y + 4x +2= 0

We can rearrange this equation by subtracting 4x \text{ and } 2 from both sides:

y = -4x - 2

So, the y-intercept is -2, and the gradient is -4x – so each time x increases by 1, y increases by 4

The result should look,

### Worksheets and Exam Questions

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

Level 1-3#### Straight Line (2)

Level 1-3### Videos

#### Drawing Straight Line Graphs Q1

GCSE MATHS#### Drawing Straight Line Graphs Q2

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