IB Maths AI SL Topic 2 — Linear Functions Paper 1 & 2 Gradient rules ~6 min read

Parallel & Perpendicular Lines

Two simple gradient rules tell you exactly how a pair of lines relate: same gradient means parallel; gradients that multiply to −1 means perpendicular.

📘 What you need to know

The two gradient rules

Both rules — learn together Parallel:  m1 = m2    •    Perpendicular:  m1 × m2 = −1
Two lines — parallel or perpendicular? PARALLEL x y y = 2x + 1 y = 2x − 4 m1 = m2 = 2 PERPENDICULAR x y m1 = 2/3 m2 = −3/2 m1 × m2 = −1 (2/3)(−3/2) = −1
Parallel: gradients match → lines stay equidistant. Perpendicular: gradients multiply to −1 → lines cross at exactly 90°.

Finding the perpendicular gradient

Given any non-zero gradient m1, the perpendicular gradient is m2 = −1/m1. In practice: flip the fraction, then change the sign.

Quick conversions: m = 3 → perp = −1/3   •   m = −2/5 → perp = 5/2   •   m = 1 → perp = −1   •   m = 0 → perp is vertical (undefined).

🧭 Recipe — parallel / perpendicular problems

  1. Rearrange the given line into y = mx + c form to read its gradient.
  2. Identify the required gradient: parallel → same; perpendicular → −1 ÷ given.
  3. Use the point given with point-gradient form: yy1 = m(xx1).
  4. Rearrange into the form the question asks for.
  5. For “are they perp?” questions: multiply the two gradients; if = −1, yes; otherwise no.

Worked examples

WE 1

Parallel line through a point

A line passes through (3, 5) and is parallel to y = 2x + 7. Find its equation in the form y = mx + c.

Parallel ⇒ same gradient m = 2 Use point-gradient with (3, 5) y − 5 = 2(x − 3) y − 5 = 2x − 6 y = 2x − 1 check at (3, 5): 2(3) − 1 = 5 ✓
WE 2

Perpendicular line through a point

A line passes through (4, −2) and is perpendicular to y = 3x + 1. Find its equation in the form y = mx + c.

Perpendicular ⇒ m_perp × 3 = −1 m_perp = −1/3 Point-gradient with (4, −2) y − (−2) = (−1/3)(x − 4) y + 2 = −x/3 + 4/3 y = −x/3 + 4/3 − 2 y = −(1/3)x − 2/3 check at (4, −2): −(1/3)(4) − 2/3 = −4/3 − 2/3 = −2 ✓
WE 3

Are these lines parallel?

Determine whether the lines y = 4x − 3 and 8x − 2y + 5 = 0 are parallel.

Line A is already in y = mx + c form m_A = 4 Rearrange Line B into y = mx + c 8x − 2y + 5 = 0 −2y = −8x − 5 y = 4x + 5/2 m_B = 4 Compare gradients m_A = m_B = 4 yes, the lines are parallel they have different y-intercepts (−3 and 5/2) so they’re distinct lines, not the same one.
WE 4

Are these lines perpendicular?

Determine whether the lines 2x + 3y = 6 and 3x − 2y + 4 = 0 are perpendicular. Justify your answer.

Rearrange both lines to y = mx + c A: 3y = −2x + 6 → y = −(2/3)x + 2 m_A = −2/3 B: −2y = −3x − 4 → y = (3/2)x + 2 m_B = 3/2 Multiply gradients m_A × m_B = (−2/3) × (3/2) = −1 yes, perpendicular (because m₁ × m₂ = −1)
WE 5

Perpendicular bisector of a segment

A is the point (2, 1) and B is the point (8, 9). Find the equation of the perpendicular bisector of AB in the form ax + by + d = 0 with integer coefficients.

Midpoint of AB M = ((2+8)/2, (1+9)/2) = (5, 5) Gradient of AB m_AB = (9 − 1)/(8 − 2) = 8/6 = 4/3 Perpendicular gradient m_perp = −3/4 Point-gradient through (5, 5) y − 5 = (−3/4)(x − 5) 4y − 20 = −3x + 15 3x + 4y − 35 = 0 check midpoint (5, 5): 3(5) + 4(5) − 35 = 15 + 20 − 35 = 0 ✓
WE 6

Find an unknown coefficient

The lines 3x + ky − 7 = 0 and 2x − 5y + 11 = 0 are perpendicular. Find the value of k.

Find each gradient (in terms of k for line 1) L_1: ky = −3x + 7 → m_1 = −3/k L_2: 5y = 2x + 11 → m_2 = 2/5 Perpendicular ⇒ m_1 × m_2 = −1 (−3/k) × (2/5) = −1 −6/(5k) = −1 6 = 5k k = 6/5 check: m_1 = −3/(6/5) = −5/2; (−5/2) × (2/5) = −1 ✓

💡 Top tips

⚠ Common mistakes

That completes Linear Functions & Graphs. Up next per the syllabus: more function families (quadratics, exponentials, piecewise) and how to use them as models.

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