Gradients of straight-line graphs The gradient of a line is a measure of how steep the line is. an upwards slope a horizontal line a downwards slope y x y y x x Positive gradient Zero gradient Negative gradient If a line is vertical, its gradient cannot be specified.

Finding the gradient from two given points If we are given any two points (x1, y1) and (x2, y2) on a line we can calculate the gradient of the line as follows: y the gradient = change in y change in x (x2, y2) y2 – y1 Draw a right-angled triangle between the two points on the line as follows. (x1, y1) x2 – x1 Teacher notes Explain how drawing a right-angled triangle on the line helps us calculate its gradient as shown in the previous activity. Explain too that since, for a straight line, the change in y is proportional to the corresponding change in x, the gradient will be the same no matter which two points we choose on a line. This will even be true if x2 < x1. y2 – y1 Gradient = x x2 – x1

Calculating gradients Teacher notes By looking at a variety of examples, establish that if we are given the coordinates of any two points on a line we can find its gradient by taking the y-coordinate of the first point subtracted from the y-coordinate of the second point and dividing it by the x-coordinate of the first point subtracted from the x-coordinate of the second point. Demonstrate to pupils that when the line slopes downwards the change in the vertical distance is negative as we move from left to right and so the gradient is negative. Show that choosing different points on the same line will give the same gradient using equivalent fractions. For example, if the vertical distance between the end points is 6 and the horizontal distance between the end points is 4, the gradient of the line is 3/2. If we change the vertical distance to 9 and the horizontal distance to 6, the gradient of the line is still 3/2. Point out that it is often most useful to leave gradients as improper fractions. For example, the gradient 3/2 tells us that for every 2 squares we move along we move 3 up. Hide the value of the gradient and ask pupils to tell you the gradients of given lines. Include gradients that need to be cancelled down and negative gradients. Ask volunteers to come to the board and vary the points to give lines of specified gradients. For example, if you ask the volunteer to make a line with a gradient of –2 they could have the horizontal distance between the end points as 5 and the vertical distance between the end points as –10.

Investigating linear graphs Teacher notes Use this activity to explore the effect of changing the value of the coefficient of x (m) and the value of the number that is added on (c). Establish that m is the gradient of the line and c is the intercept on the y-axis. You may change the graph either by dragging the dots, or changing the values of m or c.

The general equation of a straight line The general equation of a straight line can be written as: y = mx + c The value of m tells us the gradient of the line. The value of c tells us where the line crosses the y-axis. This is called the y-intercept and it has the coordinate (0, c). Teacher notes Explain that the equation of a straight line can always be arranged to be in the form y = mx + c. It is often useful to have the equation of a line in this form because it tells us the gradient of the line and where it cuts the x-axis. These two facts alone can enable us to draw the line without having to set up a table of values. Ask pupils what they can deduce about two graphs that have the same value for m. Establish that if they have the same value for m, they will have the same gradient and will therefore be parallel. For example, the line y = 3x + 4 has a gradient of 3 and crosses the y-axis at the point (0, 4).

The gradient and the y-intercept

Rearranging to y = mx + c Sometimes the equation of a straight line graph is not given in the form y = mx + c. The equation of a straight line is 2y + x = 4. Find the gradient and the y-intercept of the line. Rearrange the equation by performing the same operations on both sides. 2y + x = 4 subtract x from both sides: Teacher notes Explain that if the equation of a line is linear (that is if x and y are not raised to any power except 1), it can be arranged to be in the form y = mx + c. Explain that the equation of a (straight) line can always be arranged to be in the form y = mx + c. (This is not true for lines parallel to the y-axis.) It is often useful to have the equation of a line in this form because it tells us the gradient of the line and where it cuts the y-axis. These two facts alone can enable us to draw the line without having to draw up a table of values. 2y = –x + 4 y = –x + 4 2 divide both sides by 2: y = – x + 2 1 2

Rearranging to y = mx + c Once the equation is in the form y = mx + c we can determine the value of the gradient and the y-intercept. y = – x + 2 1 2 1 2 – m = so the gradient of the line is . c = 2 and so the y-intercept is (0, 2). x y y = – x + 2 1 2

Substituting values into equations A line with the equation y = mx + 5 passes through the point (3, 11). What is the value of m? To solve this problem we can substitute x = 3 and y = 11 into the equation y = mx + 5. This gives us: 11 = 3m + 5 subtract 5 from both sides: 6 = 3m Teacher notes Discuss ways to solve the problem. Some pupils may suggest plotting the point (3, 11) and drawing a straight line through this and the point (0, 5). The gradient of the resulting line will give the value for m. Ask pupils if they can suggest a method that does not involve drawing a graph. Establish that if the line passes through the point (3, 11) then we can substitute x = 3 and y = 11 into the equation y = mx + 5. Reveal the equation 11 = 3m + 5 on the board and talk through the steps leading to the solution of this equation. divide both sides by 3: 2 = m m = 2 The equation of the line is therefore y = 2x + 5.

What is the equation of the line?

Match the equations to the graphs