Find m, then substitute a point to find c: from a gradient and a point, from two points, or from a graph.
What you will learn
- Find the equation of a line from its gradient and one point
- Find the equation of a line through two points
- Find the equation of a drawn line
- Rearrange an equation into y = mx + c
Find m, then find c
Every method ends the same way: once you know the gradient m, substitute one point into y = mx + c and solve for c.
- gradient 3, through (2, 1)
- y = 3x + c
- 1 = 3 × 2 + c
- 1 = 6 + c, so c = −5
- y = 3x − 5
Check with the point: 3 × 2 − 5 = 1. It works.
The rest of this lesson and its 7 practice questions are locked.
In this lesson
- Find m, then find c
- From two points or a graph
- The slip with c
- Try it: Fit the line to two points
7 practice questions, marked as you go; about 25 minutes in all.
In this unit: Straight-line graphs
Coordinates and midpoints, gradients and intercepts, the equation of a line, parallel and perpendicular lines, and straight-line graphs in real life.
- Coordinates, midpoints and plotting lines
- y = mx + c
- Finding the equation of a line
- Parallel and perpendicular lines
- Real-life graphs