The big idea: If you know m and c, the line is immediate: y = mx + c. Put m on x. Put c at the end with its sign.
- gradient — reads directly from the coefficient of x
- y-intercept — the constant term; can be positive, negative, or zero
Writing the equation from m and c
A line has gradient −3 and y-intercept 8. Write its equation.
Step by step
- Substitute m = −3 and c = 8 directly.
Final answer
y = −3x + 8
| m | c | Equation |
|---|---|---|
| 2 | 5 | y = 2x + 5 |
| −3 | 1 | y = −3x + 1 |
| 0.5 | −4 | y = 0.5x − 4 |
| 0 | 7 | y = 7 (horizontal line) |
IB expects this form by default: Unless the question asks for a specific form, always give your final answer as y = mx + c. This is what IB expects and what the markscheme checks first.
The big idea: You have the gradient m and one point on the line. Plug those coordinates into y = mx + c and solve for c — then write the full equation.
Worked example — gradient and one point
Find the equation of the line with gradient 3 that passes through (2, 7).
Step by step
- Start with y = mx + c using m = 3.
- Substitute the point (2, 7): let x = 2 and y = 7.
- Solve for c.
- Write the full equation.
Final answer
y = 3x + 1
The c ≠ y-value trap: Students often write c = 7 because that is the y-coordinate of the given point. But c is the y-intercept (at x = 0), not the y-value at the given point. Always solve for c properly.
Worked example 2 — negative gradient
Find the equation of the line with gradient −2 that passes through (3, 4).
Step by step
- Write y = −2x + c.
- Substitute (3, 4).
- Solve for c.
- Full equation.
Final answer
y = −2x + 10
IB marks this in two steps: Step mark for correct substitution into y = mx + c (or equivalent method). Answer mark for the correct full equation. Show both steps clearly.
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The big idea: Two points are enough to uniquely define a straight line. Find the gradient first (using the gradient formula), then find c using either point.
Two-step method
- Step 1 — Use m = (y₂−y₁)/(x₂−x₁) to find the gradient.
- Step 2 — Substitute m and one point into y = mx + c, then solve for c.
Worked example — two points
Find the equation of the line through (1, 2) and (4, 11).
Step by step
- Find the gradient.
- Substitute m = 3 and point (1, 2).
- Write the equation.
- Verify with the second point (4, 11).
Final answer
y = 3x − 1
Always verify: Substitute the second point into your final equation to check. If it does not give the correct y-value, you made an error — find it before moving on.
Shortcut when one point is on the y-axis: If one of the two points sits on the y-axis, its x-value is 0. Write it as (0, k) — here k is just its y-value. Because x = 0, that y-value is the y-intercept directly: c = k. No substitution needed.__LINEBREAK__Example: given (0, 3) and (2, 9), c = 3 straight away. Find m = (9 − 3) / (2 − 0) = 3, then write y = 3x + 3.
The big idea: IB sometimes gives you a line in general form: ax + by + d = 0. This is just y = mx + c rearranged. To find m and c, rearrange for y.
Converting general form to slope-intercept
Find the gradient and y-intercept of the line 2x + 3y − 6 = 0.
Step by step
- Move 2x and −6 to the right side.
- Divide every term by 3.
Final answer
Gradient m = −2/3 and y-intercept c = 2.
Converting slope-intercept to general form
Write y = (3/2)x − 4 in the form ax + by + d = 0 with integer coefficients.
Step by step
- Multiply every term by 2 to clear the fraction.
- Rearrange so everything is on the left.
Final answer
3x − 2y − 8 = 0
Read the question carefully: If IB asks for 'the form ax + by = c', rearrange your slope-intercept equation to match. If IB says 'find the equation', use y = mx + c unless another form is specified.