0580 · 3.5Equations of linear graphs

Gradient and y-intercept form

0:00A straight line contains infinitely many points, yet one short equation can describe all of them. The equation tells us exactly which coordinates belong on that line. In this video, we will connect a line’s position and steepness to its equation. You will also learn to recognise the same line written in different forms. We need coordinates, gradient and a little algebra. A point’s x coordinate gives its horizontal position, while its y coordinate gives its vertical position.

0:26Gradient is the change in y divided by the corresponding change in x. When rearranging an equation, we perform the same operation on both sides to keep the equality true. An equation of a line is a rule satisfied by every point on that line. For example, consider y = 2x + 1. When x = 0, y = 1. Substituting x equal to two gives y equal to five.

0:49These points belong on the graph because their coordinates make the equation true. The two in this equation controls how quickly y changes as x increases. Increasing x by one increases y by two, so the gradient is two. The extra one fixes where the line crosses the y axis. This crossing is called the y intercept, and its value here is one. The general form is y equals m times x plus c.

1:13The letter m represents the gradient, and c represents the y intercept. On the y axis, x = 0, so the multiplication term disappears. That leaves y = c, which explains why the constant gives the intercept rather than the gradient. We can also understand why this form describes a straight line. Start at the point where x = 0 and y = c. A horizontal change of x produces a vertical change of m × x.

1:37Adding that change to the starting height gives y equals m times x plus c. A constant gradient keeps the graph straight.

Finding an equation from a graph

1:45Our first exam-style example asks us to find the equation of the line shown. It crosses the y axis at negative one, so c = −1. Choose the points with coordinates zero comma negative one and six comma three to calculate the gradient. These clearly marked points give a large triangle that is easy to read. The change in y is 3 − −1, which equals four. Across the triangle, x changes by six.

2:10Gradient is therefore 4/6, which simplifies to 2/3. Substituting the gradient and intercept gives y equals two over three times x minus one. We can check this using another point on the graph. At x equal to three, the equation gives two over three multiplied by three minus one. The multiplication gives two, and subtracting one gives one. This matches the plotted point with x equal to three and y equal to one.

2:36The fraction 2/3 also has a useful visual meaning. Moving three units to the right takes us two units upwards along the line. Use coordinate values, not simply the number of grid squares, because the axis scales might differ. Always calculate both changes using the same order of points.

Rearranging and finding line equations

2:51Our second example asks for the gradient and y intercept of 5x + 4y = 8. This equation is not yet in a form where we can read them directly. Subtract 5x from both sides to obtain 4y = 8 − 5x. Divide both sides by four to isolate y. This gives y = 8 − 5x, all divided by four. Dividing each term gives y equals two minus five over four times x.

3:16Reordering gives y equals negative five over four times x plus two. The gradient is −5/4, and the y intercept is two. The negative gradient tells us the line slopes downwards as we move to the right. The y intercept tells us it crosses the y axis two units above the origin. The original constant eight was not the y intercept because y still had a coefficient of four. Rearranging revealed what those numbers actually mean.

3:42A common exam mistake is dividing only one term when isolating y. Every term on the other side must be divided by four in this example. Another mistake is reading five as the gradient from the original equation. The coefficient of x gives the gradient directly only when y is isolated with coefficient one. Our third example gives two points rather than a complete graph. Find the equation through negative two comma seven and four comma negative two.

4:07The change in y is −2 − 7, giving negative nine. Across, 4 − −2 gives six. Dividing gives −9/6, which simplifies to −3/2. We now know that y equals negative three over two times x plus c. Substitute the point with x equal to negative two and y equal to seven. This gives seven equals negative three over two multiplied by negative two plus c. The multiplication simplifies to three, so 7 = 3 + c.

4:38Subtracting three gives c equal to four. The equation is therefore y equals negative three over two times x plus four. Check the other point by substituting x equal to four. This gives −6 + 4, which equals negative two. The equation correctly produces the second point’s y coordinate, supporting our result. Suppose the question requests the form a times x plus b times y equals c. Multiply our equation by two to obtain 2y = −3x + 8.

5:07Add 3x to both sides to give 3x + 2y = 8. This is the same line, expressed in a different form. Cambridge expects equations in fully simplified form. Simplify numerical fractions and combine any terms that can be collected. Follow a requested form when the question specifies one. Without a specified form, an equivalent simplified equation is suitable. Changing the arrangement does not change which points satisfy the equation.

Horizontal and vertical lines

5:33Horizontal and vertical lines need careful interpretation. On the horizontal line shown, every point has y equal to four, while x can vary. Its equation is y = 4, and its gradient is zero because there is no vertical change. This fits y equals m times x plus c with m equal to zero. On the vertical line shown, every point has x equal to negative three, while y can vary.

5:56Its equation is x = −3. Its gradient is undefined because a gradient calculation would require division by zero. We cannot write this vertical line in the form y equals m times x plus c. Do not confuse x = −3 with y = −3. The first fixes the horizontal coordinate and gives a vertical line, while the second fixes y and gives a horizontal line. A vertical line does not have zero gradient, and the line shown does not cross the y axis.

6:25These methods must work without a calculator on Paper two, and they are also useful on calculator Paper four. Keep exact fractions for gradients and coefficients rather than rounding them to decimals. If a question requests a numerical approximation, follow its accuracy instruction. Otherwise, general numerical answers that are not exact normally use three significant figures, without rounding intermediate working. The central idea is that a line’s equation describes all its points.

6:50For a line that is not vertical, gradient controls the change in height and the y intercept fixes its position. Read both directly from y equals m times x plus c, or rearrange first. For a vertical line, identify its fixed x coordinate and write x equals that value.