Equations of linear graphs.
Connect a line’s direction and position to a fully simplified equation.
Interpret y = mx + c · obtain equations from points or graphs · use different line forms
Read y = mx + c
A straight-line equation is a compact description of its direction and position. In y = mx + c, m tells us the vertical change for each unit right, while c gives the output when x is zero.
Starting at the y-intercept, repeatedly take the rise-and-run step named by m. This constructs the line and explains why two lines can have the same gradient but different intercepts. Learn the roles of m and c before reading them from more complicated forms.
m is the gradient and c is the y-intercept. The intercept point is (0, c). Both values include their signs.
Gradient = 1. y-intercept = 2. The point (0, 2) is always on this line.
Find the gradient and y-intercept of 5x + 4y = 8.
y = −54 x + 2
Gradient = −54; y-intercept = 2, at (0, 2).
Pause and explainFor y = 3x − 2, where does the line meet the y-axis?
(0, −2). Setting x to zero leaves y = −2.
Find an equation from a point and gradient
Knowing a gradient fixes a line’s direction, but many lines share that direction. One known point then fixes which of those parallel possibilities you need.
Write y = mx + c with the known m, substitute the point’s x and y values, and solve for c. You can instead use the point–gradient form, but still simplify the final equation and check that the known point satisfies it.
Find the line with gradient 3 through (2, −1).
Start with y = 3x + c, then substitute the known point.
y = 3x − 7
Point–gradient form
This is another route to the same result. For the example, y + 1 = 3(x − 2), which simplifies to y = 3x − 7.
Find the line through (−1, 4) and (3, −4).
Gradient = (−4 − 4) ÷ (3 − (−1)) = −2.
4 = −2(−1) + c, so c = 2
y = −2x + 2
When a graph is given, read two accurate points on it and follow the same method.
Pause and explainWhy is a gradient alone insufficient to determine one line?
The line can be shifted up or down without changing its gradient. A point or intercept is needed to fix its position.
Recognise equivalent forms and special cases
The same line can be described in several algebraic forms. Rearranging an equation changes its appearance, but not its solutions or its graph.
To read gradient and y-intercept, isolate y first. Treat x = k and y = k separately: the first fixes the horizontal coordinate and is vertical; the second fixes the vertical coordinate and is horizontal. Neither needs to be forced into an unsuitable pattern.
A line can be written as ax + by = c or y = mx + c when it is not vertical. Give a fully simplified equation, using the form requested.
A vertical line through (4, −2) is x = 4. A horizontal line through that point is y = −2.
Pause and explainIs x = 4 a horizontal or vertical line?
Vertical. Every point on it has horizontal coordinate 4, while its y-coordinate can vary.
Your turn. Check a known point.
Decide what information is given: two points, a gradient and point, an equation, or a graph. Extract a gradient and one point before building an equation.
Simplify the final equation as requested, then substitute a known point into it. If identifying a gradient from ax + by = c, rearrange carefully before reading the coefficient of x.
Find the gradient and y-intercept of 3x − 2y = 10.
Reveal worked answer
Find the equation through (1, 5) with gradient −2.
Reveal worked answer
Find the line through (−2, 1) and (2, 3), in the form ax + by = c with integer coefficients.