IB Maths AA SL Paper 1 & 2 12 min read

Parallel & Perpendicular Lines

Two lines can have a special relationship: parallel means they never meet, perpendicular means they cross at a right angle. Both relationships come down to one thing — the connection between their gradients.

📘

What you need to know

  • Parallel lines have the same gradient: m1 = m2
  • Perpendicular lines have gradients whose product is −1: m1 × m2 = −1
  • Perpendicular gradients are negative reciprocals — flip the fraction and change the sign
  • To check the relationship: rearrange both equations into y = mx + c form, then compare the gradients
  • Horizontal lines (y = q) and vertical lines (x = p) are perpendicular to each other

Parallel Lines

Parallel lines are equidistant — they always stay the same distance apart and never cross. The reason is simple: they have identical gradients.

If m1 = m2, then the lines are parallel

How to check if two lines are parallel

  1. Rearrange both equations into y = mx + c form.
  2. Read off the coefficient of x from each — that’s the gradient.
  3. If the two gradients are equal, the lines are parallel.

Quick spot: if both equations are already in y = mx + c form, you can compare gradients straight away — no rearranging needed.

Perpendicular Lines

Perpendicular lines cross at a right angle (90°). Their gradients are negative reciprocals of each other, which means their product is exactly −1.

If m1 × m2 = −1, then the lines are perpendicular

What is a “negative reciprocal”?

To find a perpendicular gradient, do two things: flip the fraction, then change the sign.

Finding the perpendicular gradient
Original
23
Step 1: Flip
32
Step 2: Sign
32
Check: 23 × (−32) = −1 ✓

Quick examples of perpendicular gradients

m = 4
→ perpendicular →
m = −14
m = 13
→ perpendicular →
m = −3
m = −52
→ perpendicular →
m = 25

Side-by-Side Comparison

Parallel Lines
l₁ l₂
m1 = m2

Same gradient — never intersect.

Perpendicular Lines
l₁ l₂
m1 × m2 = −1

Cross at a right angle.

Watch out for horizontal & vertical lines: the rule m1 × m2 = −1 doesn’t work here because vertical lines have an undefined gradient. But x = p and y = q ARE perpendicular — just by definition.

Worked Examples

Example 1 — Are these lines parallel?

Are the lines y = 3x + 2 and 6x − 2y + 5 = 0 parallel?

Answer:

Step 1: rearrange the second line into y = mx + c form. 6x − 2y + 5 = 0 2y = 6x + 5 y = 3x + 5/2 Step 2: compare the gradients. m₁ = 3 and m₂ = 3 Yes, parallel — both have m = 3 The y-intercepts differ, so they are not the same line.

Example 2 — Equation of a parallel line

The line l passes through the point (4, −1) and is parallel to the line with equation 2x − 5y = 3. Find the equation of l in the form y = mx + c.

Answer:

Step 1: rearrange the given line to find its gradient. 2x − 5y = 3 5y = 2x − 3 y = (2/5)x − 3/5 gradient = 2/5 Step 2: parallel ⇒ same gradient. So m = 2/5. Step 3: use point-gradient form with (4, −1). y − (−1) = (2/5)(x − 4) y + 1 = (2/5)x − 8/5 y = (2/5)x − 8/5 − 1 y = (2/5)x − 8/5 − 5/5 y = (2/5)x − 13/5

Example 3 — Find the perpendicular gradient

A line has gradient 34. Find the gradient of any line perpendicular to it.

Answer:

Step 1: flip the fraction. −3/4 flipped → −4/3 Step 2: change the sign. −4/3 → +4/3 m_perp = 4/3 Check: (−3/4) × (4/3) = −12/12 = −1 ✓

Example 4 — Are these lines perpendicular?

The line l1 has equation 3x − 5y = 7. The line l2 has equation y = 1453x. Determine whether l1 and l2 are perpendicular.

Answer:

Step 1: rearrange l₁ into y = mx + c. 3x − 5y = 7 5y = 3x − 7 y = (3/5)x − 7/5 m₁ = 3/5 Step 2: read m₂ from l₂ (already in y = mx + c). m₂ = −5/3 Step 3: multiply the gradients. m₁ × m₂ = (3/5) × (−5/3) = −15/15 = −1 Yes, perpendicular — m₁ × m₂ = −1

Example 5 — Equation of a perpendicular line

Find the equation of the line that is perpendicular to y = 2x + 1 and passes through the point (4, 3). Give your answer in the form y = mx + c.

Answer:

Step 1: read the original gradient. m₁ = 2 Step 2: find the perpendicular gradient (negative reciprocal). flip: 1/2 → change sign: −1/2 m₂ = −1/2 Step 3: use point-gradient form with (4, 3). y − 3 = −(1/2)(x − 4) y − 3 = −(1/2)x + 2 y = −(1/2)x + 2 + 3 y = −(1/2)x + 5 Check: 2 × (−1/2) = −1 ✓ — perpendicular confirmed.
💡

Tips

  • Always rearrange first. If the equation isn’t in y = mx + c form, rearrange before doing anything else.
  • Quick check trick: if you suspect two lines are perpendicular, just multiply the gradients. If you get exactly −1, you’re right.
  • For perpendicular gradients, remember the two-step rule: flip and change the sign. Don’t just flip OR change the sign — you must do both.
  • Whole numbers count too: a gradient of 4 is really 41. Flip to 14, change sign → −14.
  • State your reason: when asked to “determine and explain”, always finish with “since m1 = m2” or “since m1 × m2 = −1″ — examiners look for this.

Common mistakes

  • Forgetting to change the sign when finding a perpendicular gradient. 23 flipped is 32 — but the perpendicular gradient is 32, not 32.
  • Confusing parallel and perpendicular rules. Parallel = same gradient. Perpendicular = product is −1. Mixing them up is the most common slip.
  • Comparing gradients before rearranging. You can’t read the gradient from 2x − 5y = 3 directly — you must isolate y first.
  • Reciprocal ≠ negative reciprocal. The reciprocal of 2 is 12; the perpendicular gradient is −12. Don’t drop the minus.
  • Sign errors in fractions: watch out when the original gradient is already negative — the perpendicular gradient ends up positive.
  • Stopping at the gradient. If the question asks for the equation, you still need to use point-gradient form and rearrange to the requested form.

Final word: Two simple rules — m1 = m2 for parallel, m1 × m2 = −1 for perpendicular — unlock every parallel/perpendicular question in IB. Get these in your head and you’re sorted.

Need help with Linear Functions?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →