Straight-line graphs | OCR GCSE Maths Notes

Visual model

Gradient is rise divided by run

runrise

\text{gradient}=\frac{\text{rise}}{\text{run}}

Gold-standard guide

20 mins

What you will learn

Use gradients, intercepts and equations of lines.

Use a clear step-by-step method for straight-line graphs.

Check your answer and avoid the most common exam mistake.

Useful before you start

Core number skillsEarlier algebra skillsShowing clear working

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

y = mx + c

Step 1

Calculate the gradient

m = (−3 − 5)/(3 − (−1)) = − $\frac{8}{4}$ = −2

Step 2

Substitute one point and m into y = mx + c to find c

Using (3, −3): −3 = −2(3) + c, so −3 = −6 + c, so c = 3

Step 3

Write the equation

y = −2x + 3

Watch out

Students subtract coordinates in the wrong order, for example using (y₁ − y₂)/(x₂ − x₁), which gives the wrong sign for the gradient

Gradient

$gradient = change in \frac{y}{change} in x.$

Line equation

y = mx + c, where m is gradient and c is y-intercept.

Worked example

Find the equation of the line passing through (−1, 5) and (3, −3)

Calculate the gradient: m = (−3 − 5)/(3 − (−1)) = − $\frac{8}{4}$ = −2.The negative gradient means the line slopes downward from left to right.

Substitute one point and m into y = mx + c to find c: Using (3, −3): −3 = −2(3) + c, so −3 = −6 + c, so c = 3.

Write the equation: y = −2x + 3.

Verify with the other point: At x = −1: y = −2(−1) + 3 = 5. ✓

Final answer

y = −2x + 3

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Workedreasoning

Find the equation of the line passing through (−1, 5) and (3, −3)

4 marks4 minsstraight-line-graphs-worked

Show solution

Worked solution

1.Calculate the gradient: m = (−3 − 5)/(3 − (−1)) = − $\frac{8}{4}$ = −2.The negative gradient means the line slopes downward from left to right.
2.Substitute one point and m into y = mx + c to find c: Using (3, −3): −3 = −2(3) + c, so −3 = −6 + c, so c = 3.
3.Write the equation: y = −2x + 3.
4.Verify with the other point: At x = −1: y = −2(−1) + 3 = 5. ✓

Final answer

y = −2x + 3

Mark points

M1: calculate the gradient
M1: substitute one point and m into y = mx + c to find c
M1: write the equation
M1: verify with the other point
A1: y = −2x + 3

Watch out

g. using (y₁ − y₂)/(x₂ − x₁), which gives the wrong sign for the gradient.Always be consistent: (y₂ − y₁)/(x₂ − x₁) where you use the SAME point for subscript 1 and the SAME point for subscript 2 in both numerator and denominator.

Diagnosticrecall

Find the gradient of the line joining (2, 3) and (6, 11)

1 mark2 minsstraight-line-graphs-q1

Show solution

Worked solution

1.Spot the skill: y = mx + c.
2.Use the calculate the gradient stage first, then substitute one point and m into y = mx + c to find c.
3.Keep the final answer visible: 2.

Final answer

Mark points

M1: use the correct y = mx + c.find m (gradient) first using m = (y₂ − y₁)/(x₂ − x₁), then substitute one point to find c.the gradient tells you how steep the line is and which direction it slopes.
A1: 2

Watch out

Easyprocedure

Write the equation of a line with gradient −3 and y-intercept 4

2 marks3 minsstraight-line-graphs-q2

Show solution

Worked solution

1.Spot the skill: y = mx + c.
2.Use the substitute one point and m into y = mx + c to find c stage first, then write the equation.
3.Keep the final answer visible: y = −3x + 4.

Final answer

y = −3x + 4

Mark points

M1: use the correct y = mx + c.find m (gradient) first using m = (y₂ − y₁)/(x₂ − x₁), then substitute one point to find c.the gradient tells you how steep the line is and which direction it slopes.
A1: y = −3x + 4

Watch out

Mediumreasoning

Find the equation of the line through (0, −2) with gradient 5

3 marks4 minsstraight-line-graphs-q3

Show solution

Worked solution

1.Spot the skill: y = mx + c.
2.Use the write the equation stage first, then verify with the other point.
3.Keep the final answer visible: y = 5x − 2.

Final answer

y = 5x − 2

Mark points

M1: use the correct y = mx + c.find m (gradient) first using m = (y₂ − y₁)/(x₂ − x₁), then substitute one point to find c.the gradient tells you how steep the line is and which direction it slopes.
A1: y = 5x − 2

Watch out

Hardproblem solving

Find the equation of the line perpendicular to y = 2x + 1 that passes through (4, 3)

3 marks5 minsstraight-line-graphs-q4

Show solution

Worked solution

1.Spot the skill: y = mx + c.
2.Use the verify with the other point stage first, then calculate the gradient.
3.Keep the final answer visible: y = −x/2 + 5.

Final answer

y = −x/2 + 5

Mark points

M1: use the correct y = mx + c.find m (gradient) first using m = (y₂ − y₁)/(x₂ − x₁), then substitute one point to find c.the gradient tells you how steep the line is and which direction it slopes.
A1: y = −x/2 + 5

Watch out

Exam-stylemulti-step

Line L has equation 3x + 2y = 12. Find the gradient and y-intercept of L.

4 marks6 minsstraight-line-graphs-q5

Show solution

Worked solution

1.Spot the skill: y = mx + c.
2.Use the calculate the gradient stage first, then substitute one point and m into y = mx + c to find c.
3.Keep the final answer visible: Gradient = − $\frac{3}{2}$ , y-intercept = 6.

Final answer

Gradient = − $\frac{3}{2}$ , y-intercept = 6

Mark points

M1: use the correct y = mx + c.find m (gradient) first using m = (y₂ − y₁)/(x₂ − x₁), then substitute one point to find c.the gradient tells you how steep the line is and which direction it slopes.
A1: Gradient = − $\frac{3}{2}$ , y-intercept = 6

Watch out

Grade 9 stretchproblem solving

Find the equation of the line perpendicular to y = $\frac{1}{2}$ x + 4 that passes through (3, 1).

4 marks7 minsline-g9

Show solution

Worked solution

1.Use the negative reciprocal gradient.
2.Substitute the point into y = mx + c.
3.Solve for the intercept.

Final answer

y = -2x + 7

Mark points

M1: use gradient -2
M1: 1 = -2(3) + c
A1: y = -2x + 7

Watch out

Do not rush straight into arithmetic. Select the relevant method and show a complete chain of working.

Timed checkpoint

12 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Straight-line graphs - 2 marksFind the gradient of the line joining (2, 3) and (6, 11)Mark answer

Answer

2Algebraic notation and substitution - 2 marksFind t² + 4t when t = −2Mark answer

Answer

−4

3Simplifying expressions - 2 marksSimplify 2xy + 3x²y − xy + x²yMark answer

Answer

3x²y + xy

4Expanding and factorising - 3 marksFactorise fully 6x² + 9xMark answer

Answer

3x(2x + 3)

Mastery check

I can explain the method for straight-line graphs.
I can show clear working without skipping key steps.
g. using (y₁ − y₂)/(x₂ − x₁), which gives the wrong sign for the gradient.Always be consistent: (y₂ − y₁)/(x₂ − x₁) where you use the SAME point for subscript 1 and the SAME point for subscript 2 in both numerator and denominator.

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