Parallel ⇒ equal gradients: Two lines are parallel when they have the same gradient (m₁ = m₂) — the same steepness, so they never meet.
Only their y-intercepts differ.
| Line A | Line B | Parallel? | Why |
|---|---|---|---|
| y = 2x + 1 | y = 2x − 5 | Yes | same gradient 2, different c |
| y = −3x + 4 | y = −3x + 4 | No — same line | identical, not parallel |
| y = 5x − 2 | y = −5x − 2 | No | gradients 5 and −5 differ |
| y = ¾x + 1 | y = ¾x − 6 | Yes | same gradient ¾ |
IB-style question — are they parallel?
Show that y = 3x + 1 and 6x − 2y + 5 = 0 are parallel.
Step by step
- Rearrange the second line into y = mx + c.
- Compare gradients.
Final answer
Both gradients are 3, so the lines are parallel.
IB-style question — find the missing coefficient
The line ax + 2y − 6 = 0 is parallel to y = 3x + 1.
Find the value of a.
Step by step
- Parallel lines share a gradient, so read the target gradient.
- Find the gradient of the first line (m = −a/b).
- Set the gradients equal and solve.
Final answer
a = −6. IB often phrases this as 'the lines are parallel — find the missing coefficient': set the gradients equal and solve.
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it free for 7 days:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
Spot it: 'parallel to … through a point': Copy the gradient, then anchor it at the given point.
If that point is the origin, the y-intercept is 0, so the answer is simply y = mx.
Part (a) — parallel through the origin
Let f(x) = 3x − 4.
The line g is parallel to f and passes through the origin.
Find an expression for g(x).
Step by step
- Parallel ⇒ same gradient as f.
- Through the origin, the y-intercept c = 0, so y = mx.
Final answer
g(x) = 3x.
Part (b) — parallel through a general point
A second line p is parallel to f and passes through (2, 5).
Find p(x).
Step by step
- Same gradient as f.
- Point–gradient form through (2, 5).
- Tidy to y = mx + c.
Final answer
p(x) = 3x − 1.
Don't change the gradient: Parallel keeps the gradient (m₁ = m₂).
Only c changes — find it by substituting the given point.