Forms of the Equation of a Straight Line
Move between y = mx + c, y - y₁ = m(x - x₁) and ax + by + c = 0, and produce the equation of a line from a point and a gradient or from two points.
What a learner can do afterwards
- Find the equation of the line through (2, 5) and (6, 13)
- Rewrite 3x + 4y - 12 = 0 in gradient-intercept form and state its gradient
- Find where a given line meets each axis
1 · Read
Last stop you turned a rule into a picture. Now the picture comes first. On the Ridgeway plan the practice strip is a straight fence line, and the crew has marked two pegs on the squared paper. They need the exact equation, not a drawing. A line's equation is a test: every point on the line makes it true, and every point off the line does not. That is why setting y = 0 finds where the line meets the x-axis, and setting x = 0 finds where it meets the y-axis. Those two crossings are often all a plan needs.
You already write a line as y = mx + c, gradient m and height c at x = 0. Two more forms earn their place. The point-gradient form y - y₁ = m(x - x₁) takes a gradient and one point (x₁, y₁), no rearranging needed. Some books say slope for gradient, so this is also point-slope form. The standard form ax + by + c = 0 clears fractions and keeps whole numbers. Only one line sits outside y = mx + c: the vertical line x = k, whose gradient is undefined. A horizontal line y = k is that form with m = 0.
Write the line through (5, 3) with gradient 4. Start with the form that takes those two pieces straight: y - y₁ = m(x - x₁), so y - 3 = 4(x - 5). That is already an answer. To reach gradient-intercept form, expand and tidy: y - 3 = 4x - 20, so y = 4x - 17. For standard form, move everything to one side: 4x - y - 17 = 0. Check the point in that last one: 4(5) - 3 - 17 = 0. All three describe the same line, and which you write depends on what the question wants next.
Now the line through the two pegs at (2, 5) and (6, 13), with no gradient given. Find it first: divide the change in y by the change in x, so m = (13 - 5) / (6 - 2) = 8 / 4 = 2. Then use either point in the point-gradient form. Taking (2, 5) gives y - 5 = 2(x - 2), which tidies to y = 2x + 1. Taking (6, 13) gives the same line. Standard form is 2x - y + 1 = 0, and a fractional gradient would need every term multiplied through first.
Standard form hides the gradient, so dig it out. Take 3x + 4y - 12 = 0. Move the x term and the number across: 4y = -3x + 12. Divide every term by 4: y = -(3/4)x + 3. So the gradient is -3/4 and the line cuts the y-axis at (0, 3). For the other crossing set y = 0 in the original: 3x - 12 = 0, so x = 4 and the line cuts the x-axis at (4, 0). Going the other way, multiply through by the denominator until every coefficient is a whole number.
One line, three ways to write it. Use y = mx + c to read a gradient and a y-intercept, y - y₁ = m(x - x₁) when you have a point and a gradient, and ax + by + c = 0 to keep whole numbers. From two points, work out the gradient first and then use either point. Set y = 0 for the x-axis crossing and x = 0 for the y-axis crossing.
2 · Watch
Take it off screen
Where it sits
Where this leads
Jobs that lean on this skill. Follow one to see everything it is built on.
24 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.