Mathematics • Year 8 • Unit 2 • Lesson 11

Equation of a Line: y = mx + c

Build fluency with gradient-intercept form. One worked example, one guided fill-in, then eight independent problems from reading m and c to rearranging into y = mx + c form.

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every line. Each step has a reason so you can see why, not just what.

Problem. A line has gradient m = 2 and y-intercept c = 3. Write its equation in y = mx + c form.

Step 1 — Recall the template.

y = mx + c

Reason: gradient-intercept form encodes any straight line using just two numbers — the gradient m and the y-intercept c.

Step 2 — Substitute m = 2 and c = 3.

y = (2)x + (3)

Reason: m is the coefficient of x; c is the constant added on the end.

Step 3 — Tidy up.

y = 2x + 3

Reason: no brackets are needed around a positive 2 or a positive 3.

Step 4 — Sanity check. At x = 0, y = 2(0) + 3 = 3. ✓ The line crosses the y-axis at (0, 3) as required.

Answer: y = 2x + 3.

Stuck? Revisit lesson § Card 1 — m sits in front of x, c sits on its own at the end.

2. We do — fill in the missing steps

Same shape as Section 1, but with the working faded. Fill in each blank. 4 marks

Problem. A line has gradient m = −1 and y-intercept c = 4. Write its equation.

Step 1 — Template: y = ____ x + ____ .

Step 2 — Substitute m = −1 and c = 4:

y = ( ____ )x + ( ____ )

Step 3 — Simplify (remember: −1 × x is just −x):

y = ______ + ______

Step 4 — Check at x = 0:

y = −(0) + 4 = ______ , so the line crosses the y-axis at (0, ______ ). ✓

Stuck? Revisit lesson § Card 2, Worked Example 2 — negative gradient: (−1)x simplifies to −x.

3. You do — independent practice

Show your working in the space under each problem. The first four are foundation (read or write). The middle two are standard (substitute carefully). The last two are extension (rearrange first).

Foundation — read or write directly

3.1 Write the equation of a line with gradient m = 4 and y-intercept c = 1.    1 mark

3.2 State the gradient m and y-intercept c for y = 5x − 2.    1 mark

3.3 Write the equation of a line with gradient m = −3 and y-intercept c = 5.    1 mark

3.4 State m and c for y = x − 4. (Hint: the coefficient of x is 1 even when it's not written.)    1 mark

Standard — substitute carefully

3.5 Write the equation of a line that has gradient ½ and passes through (0, 0).    2 marks

3.6 State m and c for y = −x + 3, then find y when x = 2. Show every substitution.    2 marks

Extension — rearrange first

3.7 Rearrange 2x + y = 5 into y = mx + c form, then state m and c.    2 marks

3.8 Rearrange 3y = 6x + 9 into y = mx + c form, then state m and c. (Hint: divide every term by 3.)    2 marks

Stuck on 3.7 / 3.8? To get y by itself, undo what's stuck to it: subtract to move terms across, divide to remove a coefficient.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (m = −1, c = 4)

Step 1: y = mx + c.
Step 2: y = (−1)x + (4).
Step 3: y = −x + 4.
Step 4: y = −(0) + 4 = 4; line crosses at (0, 4). ✓

3.1 — m = 4, c = 1

y = mx + c → y = (4)x + (1) → y = 4x + 1.

3.2 — y = 5x − 2

Coefficient of x is 5, so m = 5. Constant term is −2, so c = −2. (Watch the sign — the minus belongs to c.)

3.3 — m = −3, c = 5

y = (−3)x + (5) → y = −3x + 5.

3.4 — y = x − 4

The hidden coefficient is 1, so m = 1; constant term is −4, so c = −4.

3.5 — Gradient ½, through (0, 0)

Through the origin means y-intercept is 0, so c = 0. y = (½)x + (0) → y = ½x (or equivalently y = x/2).

3.6 — y = −x + 3 at x = 2

m = −1, c = 3. At x = 2: y = −(2) + 3 = 1. So the point (2, 1) lies on the line.

3.7 — 2x + y = 5

Subtract 2x from both sides: y = 5 − 2x. Rewrite with the x-term first: y = −2x + 5. So m = −2, c = 5.

3.8 — 3y = 6x + 9

Divide every term by 3: y = 2x + 3. So m = 2, c = 3.