Straight-line graphs are graphical representations of linear equations where the highest power is 1. Their shape is a straight line, and the gradient is constant along the line.
The equation for straight lines
All linear lines can be expressed in the format \(y= mx + b\), where:
- y is the value of the y coordinate of a point on the straight line.
- x is the value of the x-coordinate of the same point on the straight line.
- m is the gradient of a straight-line graph, which can be found using
\[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{difference in the y coordinates}}{\text{difference in the x coordinates}}\]
The gradient is defined as the steepness of the line at a given point.
- b is the value of the y coordinate when the straight line intersects with the y-axis (x = 0).
Find the straight line equation between the points (-1, 2) and (0, 8). Please leave your answer in the form \(y = mx + b\)
Let A = (-1, 2) and C = (0, 8). The gradient can be found using the equation and points
A = (-1, 2) and C = (0, 8).
\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8-2}{0-(-1)} = \frac{6}{1} = 6\)
Another method of calculating the gradient is by using the equation \(m = \frac{\text{difference in the y coordinates}}{\text{difference in the x coordinates}}\)
This equation requires the drawing of the right-angled triangle underneath the line and its coordinates, including point B = (0, 2), where the angle of the triangle is 90 °. We can see this graphically as:
Graphical representation of example. Jaime Nichols, StudySmarter Original
\[m = \frac{\Delta y}{\Delta x} = \frac{6}{1} = 6\]
As point C (0, 8) is on the y-axis, the y-coordinate becomes b in the equation.
Therefore, the equation is \(y = 6x+8\)
You could rewrite the previous example in the form \(y - y_1 = m(x-x_1)\) using the point A (-1, 2) and m = 6 as follows: \(y -2 = 6(x-(-1)) = y -2 = 6(x+1)\).
The equation can also be written in the form as \(Ax + By = C\). Unlike the first two equations, this form cannot be achieved by directly substituting the values into the formula. Instead, you must find the equation in one of the first two equations and then rearrange it into the form \(Ax + By = C\).
A straight line has a gradient of \(\frac{1}{2}\) and goes through the point of (0, 10). Write the equation for this straight-line in the form \(Ax + By = C\).
First, write the equation of the straight line in one of the first two forms, where \(m = \frac{1}{2} \text{ and } b = 10\)
\(y = \frac{1}{2}x+10\)
Next, A, B and C should be integers so you need to multiply both sides of the equation by two to remove the fraction.
\(2y = x + 20\)
Finally, you need to move the x to the other side so that it is in the form \(Ax + By = C\).
\(2y - x = 20\)
Finding coordinates through the straight line equation
You might be asked to find the coordinates using a linear equation. To do so, you substitute one of the values into the linear equation to get the other.
Line A has the linear equation of \(y = 10x - 4\). What is the y-coordinate of the point on the line when x = 14?
As you know the value of x, you can substitute it into the equation.
\(y = 10(14) - 4\)
\(y = 136\)
Therefore, the answer is (14, 136).
Line B has the linear equation of \(y = 2x + 7\). What is the x-coordinate of the point on the line when y = 17?
As you know the value of y, you can substitute it into the equation.
\(17 = 2x + 7\)
\(10 = 2x\)
\(x = 5\)
Therefore, the coordinate is (5, 17)
It is important to give your answer in the form asked in the question. If you are asked to give the coordinates, make sure you give your answer as a coordinate. This is a common mistake but is easy to avoid.
How do you plot a straight line graph?
In order to plot a straight-line graph, you need to:
Plot a table of the x values and y values.
Draw your axis on the graph paper and label your axis if they apply to a real-world situation.
Plot the given points on the graph.
Join all the points with a single straight line using a ruler.
Draw the line \(y = 2x +1\)
1. Plot a table of the x and y values
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| y | -5 | -3 | -1 | 1 | 3 | 5 | ![]() |
2. Draw your axis
Plain axis
3. Plot your points
Plotted points on the axis, Jaime Nichols, StudySmarter Originals
4. Join all the points with a straight line:
Complete plotted graph, Jaime Nichols, StudySmarter Originals
Characteristics of gradients of straight lines
Being aware of the characteristics of the gradients of straight lines will make it easier for you to calculate the more challenging equation of straight line questions. Using the previous example y = 6x + 8, we will go through the characteristics of other types of gradients.
Negative gradient slope
The graph illustrates the lines \(y = -6x + 8\) and \(y = 6x + 8\).
Negative gradient on graph, Jaime Nichols, StudySmarter Originals
We can make two observations from this graph about the negative gradient:
The y-intercept (0, 8) is the same for both the positive and negative gradient, therefore having a negative gradient has no effect on the y-intercept.
As we increase the x- variable, the y- variable decreases and therefore, the line travels in a downward diagonal direction.
The gradient of parallel lines
Parallel lines are lines that exist on the same graph but do not meet, maintaining the same distance apart continuously.
The graph below depicts two parallel lines, \(y = 6x + 8\) and \(y = 6x + 15\)
Graph of parallel lines, Jaime Nichols, StudySmarter Originals
There are two observations you can make about parallel lines:
- When two lines are parallel, the gradients are the same. In this example, both lines have a gradient of 6.
- The lines have different values for b and therefore have different y-intercepts.
The gradient of perpendicular lines
The graph below shows two perpendicular lines: \(y = 6x + 8\) and \(y = - \frac{1}{6}x +8\).
Graph of perpendicular lines, Jaime Nichols, StudySmarter Originals
There are two important observations that can be made about perpendicular line gradients.
In order to have the lines intercept with each other, the gradients of the two straight lines need to be negative inverse reciprocals of each other. A negative reciprocal of a gradient is given through the formula \(-\frac{1}{m}\), where m is the original gradient. In this example, the gradients of the two lines 6 and \(-\frac{1}{6}\) are negative reciprocals of each other.
The y-intercept is the same for both lines as the point of interception lies on the y-axis. The y-intercept would be different for each line if the line of intersection did not lie on the y-axis. For instance, if the point of intersection was (1, 14), then you would substitute it into formula: \(y = mx + b\)
\(14 = \frac{-1(1)}{6} + b\)
\(\frac{84}{6} = \frac{-1}{6} + b\)
\(b = \frac{84}{6} + \frac{1}{6} = \frac{85}{6}\)
Graph of perpendicular lines, Jaime Nichols, StudySmarter Originals
How do you find the equation of a straight-line graph from a set of points?
There are several different types of questions you can be asked, which will now go through with examples.
Finding the straight line equation where the y-intercept is unknown.
Sometimes you are not told the y-intercept which means we need to calculate a value for C if we are using the form \(Ax + By = C\). To do that we need the following steps:
- Calculate the gradient
- Substitute the values for x and y from a point on the line into the equation.
- Rearrange to get A value for B
Line C is a straight line that goes through 2 points (2, 4) and (4,7). What is the equation for Line C?
- Calculate the gradient: \(m = \frac{7-4}{4-2} = \frac{3}{2}\)
- Substitute the gradient and one of the points into the equation. We will use the point (2, 4): \(4 = \frac{3}{2}(2) + b\)
- Rearrange to get a value for b: \(4 = 3 + b \qquad b = 4-3 =1\) Therefore, the answer is \(y = \frac{3}{2}x+1\)However, you could be asked to write the equation, \(Ax + By = C\). The C in this equation is not the same as the b in the first equation, therefore we need to continue to rearrange.
\(y = \frac{3}{2}x+1\)
\(2y=3x+2\)
\(2y-3x = 2\)
If you struggle with this step and you are not told to give a specific form, it might be easier for you to work with the equation \(y-y_1 = m(x-x_1)\) as you do not need to rearrange to find a value for b. For the above example, you need to find the gradient and substitute a point into the equation: \(y -4 = \frac{3}{2}(x-2)\).
Finding the equation of a straight line from graphs
Finding the linear equation from the graph is similar to the above method, with slight differences.
Identify two points on the line
Calculate the gradient using these two points
Find the value for b by looking where the line crosses the y-intercept
Formulate the equation.
Find the linear equation for the graph below.
Triangle used to find the gradient, Jaime Nichols
- Identify two points on the line: (0, 2) and (1, 1) sit on the line.
- Find the gradient: \(m = \frac{1-2}{1-0} = \frac{-1}{1} = -1\)
- Find the y-intercept: Point (0, 2) sits on the y-axis therefore b = 2.
- Substitute values into a formula: \(y = -x +2\)
Horizontal lines
As the gradient of horizontal straight lines is 0, the equation of the line is y = b, where b is the y-intercept.
Graph of y = b, Jaime Nichols, StudySmarter Originals
This horizontal line has the equation y = 10, as it intersects the y-axis at (0, 10).
Vertical lines
The gradient of vertical lines is infinity and can be expressed in the equation x = d, whereby d is where the line intersects with the x-axis.
Graph of x = d, Jaime Nichols, StudySmarter Originals
This vertical line has the equation x = 0.5, because it intersects with the x-axis at the point (0.5, 0)
Using straight line graphs for real-life examples
Straight line graphs can be used to represent the relationships between any two variables. For example, businesses use it to show how many sales they make over time.
You can see an example below of a straight line graph demonstrating the number of total sales over time, in this example we use (y) as the total number of cars sold and (x) as the number of weeks.
Example of a straight line graph, Jaime Nichols
To find the rate of change, we need to determine the gradient. We know from the equation that the gradient is 3. This means that the rate of the number of sales is 3 cars per week.
If you are asked to contextualise the gradient in an exam, it is important to refer to it as the rate of something to get the marks.
You might also have to explain the position of the y-intercept. In this example, the line intersects with the origin which makes sense because, at 0 weeks, they haven't had started to sell any cars.
Sometimes, especially in straight-line graphs involving costs, the line might intersect above the origin. This normally suggests there were some fixed initial costs that needed to be taken into consideration. A common example used in exams is a deposit. Importantly, this cost is fixed and therefore does not change regardless of the change in x and y.
Straight line graphs - key takeaways
- \(y = mx + b\) is the main straight line equation but you can also use \(y - y_1 = m (x-x_1)\) and \(Ax + By = C\)
- In the equation for a straight line, m represents the gradient, which can be found through the formula: \(\frac{y_2-y_1}{x_2-x_1}\) and c is the y-intercept of the straight line graph
- Parallel lines have the same gradient while the gradients of perpendicular lines are the negative reciprocal of each other.
- The equation for a vertical line is x = d, where d is the intersection with the x-axis.
- The equation for a horizontal line is y = b, where b is the intersection with the y-axis.
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