IB Maths AA HLTopic 2 — FunctionsPaper 1 & 2~8 min read
Parallel & Perpendicular Lines
Now that you’ve got the three forms of a line down, here’s where the gradient really earns its keep. Two lines tell you everything about how they sit on the plane just by their gradients. Same gradient → parallel (they never meet). Gradients multiply to give −1 → perpendicular (they cross at a right angle). That’s the whole rule. Most exam questions on this topic are some variation of “find the line through this point that’s parallel/perpendicular to that one” — once you know the gradient game, the rest is just plugging into point-gradient form from the previous note.
📘 What you need to know
Parallel lines have the same gradient:m1 = m2. They never meet.
Perpendicular lines have gradients that multiply to −1:m1 × m2 = −1. The two gradients are negative reciprocals of each other.
The “negative reciprocal” trick: flip the fraction and change the sign. So gradient 2/3 → perpendicular gradient is −3/2. Gradient 4 → perpendicular gradient is −1/4.
To check parallelism or perpendicularity from given equations: rearrange both into y = mx + c first, then compare the gradients.
Special case for horizontal & vertical lines:y = k (horizontal) and x = h (vertical) are perpendicular to each other, even though the rule “m1 × m2 = −1″ doesn’t quite apply (one gradient is undefined).
Once you know the gradient of the line you want, build the equation using point-gradient form from the previous note: y − y1 = m(x − x1).
Parallel lines — same gradient, no meeting
Two lines are parallel when they go in exactly the same direction — they’re equidistant everywhere and never cross. In gradient terms:
Parallel linesm1 = m2 ⟺ lines are parallel
Just compare the gradients. If they’re equal, the lines are parallel. If the equations are not in y = mx + c form, rearrange them first.
A quick warning: same gradient AND same y-intercept means the two equations describe the same line — not two parallel lines. Two genuinely different parallel lines must have the same gradient but different y-intercepts.
Perpendicular lines — meet at 90°
Two lines are perpendicular when they cross at a right angle. The relationship between their gradients is:
Perpendicular linesm1 × m2 = −1 ⟺ lines are perpendicular
So if one gradient is m, the perpendicular gradient is −1/m. That’s the “negative reciprocal” — flip and negate.
Examples
m = 5 → perp m = −1/5
m = 2/3 → perp m = −3/2
m = −1/4 → perp m = 4
flip the fraction, swap the sign
Quick check
multiply them
if the product is −1, they’re perpendicular. Anything else (including −2 or +1) → not perpendicular.
🤔 Why does m1 × m2 = −1?
Picture rotating a line by 90° about a point. A line going up at gradient m (rise m, run 1) becomes a line where the “rise” and “run” swap roles, and one of them flips sign. So the new gradient becomes −1/m. Multiplying the old and new gradients gives m × (−1/m) = −1. The minus sign comes from the 90° rotation reversing direction.
The special case — horizontal and vertical lines
The rule “gradients multiply to −1” runs into a snag for horizontal and vertical lines:
Horizontal
y = k
gradient = 0 (no rise)
Vertical
x = h
gradient is undefined (run = 0)
You can’t multiply 0 by “undefined” and get −1, but it’s still true geometrically that horizontal and vertical lines are perpendicular. So just remember this as a special case.
Parallel: same gradient. Perpendicular: gradients multiply to −1
How to use these in exam questions
Most questions follow one of two patterns. Here’s the recipe for both:
🧭 Recipe — find a line parallel to a given line through a given point
Rearrange the given line into y = mx + c if needed, and read off its gradient.
Use the same gradient for your new line.
Plug the gradient and given point into point-gradient form: y − y1 = m(x − x1).
Rearrange to whatever form the question asks for.
🧭 Recipe — find a line perpendicular to a given line through a given point
Find the gradient of the given line.
Take the negative reciprocal — flip the fraction and change the sign.
Plug into point-gradient form with the new gradient and the given point.
Rearrange to the form the question asks for.
Worked examples
WE 1
Determine if two lines are parallel
Are the lines l1: y = 2x + 5 and l2: 4x − 2y + 3 = 0 parallel?
Step 1: Read off the gradient of l₁l₁: y = 2x + 5 → m₁ = 2Step 2: Rearrange l₂ into y = mx + c form4x − 2y + 3 = 02y = 4x + 3y = 2x + 3/2 → m₂ = 2Step 3: Compare gradientsm₁ = 2 and m₂ = 2 → equalyes, the lines are parallel ✓they have different y-intercepts (5 vs 3/2) so they’re genuinely two different parallel lines, not the same line
WE 2
Find a parallel line through a given point
Find the equation of the line passing through (1, 2) and parallel to y = −3x + 4. Give your answer in the form y = mx + c.
Step 1: Read off the gradient of the given liney = −3x + 4 → m = −3Step 2: Parallel → use the same gradientnew line: m = −3Step 3: Plug into point-gradient form using (1, 2)y − 2 = −3(x − 1)y − 2 = −3x + 3y = −3x + 5y = −3x + 5check: (1, 2) → −3(1) + 5 = 2 ✓ and the gradient matches the original line
WE 3
Find a perpendicular line through a given point
Find the equation of the line passing through (2, 5) and perpendicular to y = 12x − 7. Give your answer in the form y = mx + c.
Step 1: Read off the gradient of the given liney = (1/2)x − 7 → m₁ = 1/2Step 2: Take the negative reciprocal for the perpendicular gradientm₂ = −1/m₁ = −2check: m₁ × m₂ = (1/2)(−2) = −1 ✓Step 3: Plug into point-gradient form using (2, 5)y − 5 = −2(x − 2)y − 5 = −2x + 4y = −2x + 9y = −2x + 9check: (2, 5) → −2(2) + 9 = 5 ✓ — point lies on the line
WE 4
Determine if two lines are perpendicular
Are the lines l1: 2x + 3y = 6 and l2: 3x − 2y = 4 perpendicular? Justify your answer.
Step 1: Rearrange l₁ to find gradient2x + 3y = 6 → 3y = −2x + 6 → y = −(2/3)x + 2m₁ = −2/3Step 2: Rearrange l₂ to find gradient3x − 2y = 4 → 2y = 3x − 4 → y = (3/2)x − 2m₂ = 3/2Step 3: Multiply the gradientsm₁ × m₂ = (−2/3)(3/2) = −6/6 = −1yes, l₁ and l₂ are perpendicular because m₁ × m₂ = −1 ✓notice m₁ and m₂ are negative reciprocals: flip 2/3 and change sign → −3/2… wait, that’s the negative of m₂. They check out: −2/3 and 3/2 ARE negative reciprocals of each other.
WE 5
Perpendicular line in general form
Find the equation of the line passing through (3, 4) and perpendicular to the line 2x + y = 5. Give your answer in the form ax + by + d = 0 with integer coefficients.
Step 1: Find gradient of the given line2x + y = 5 → y = −2x + 5 → m₁ = −2Step 2: Perpendicular gradientm₂ = −1/m₁ = −1/(−2) = 1/2Step 3: Plug into point-gradient form using (3, 4)y − 4 = (1/2)(x − 3)Step 4: Multiply through by 2 to clear the fraction2(y − 4) = x − 32y − 8 = x − 3Step 5: Move everything to the left0 = x − 3 − 2y + 8x − 2y + 5 = 0x − 2y + 5 = 0 (integer coefficients ✓)always clear fractions BEFORE moving things across the equals sign — saves arithmetic mistakes
WE 6
Show a triangle is right-angled
The points A(−1, 0), B(2, 4) and C(6, 1) form a triangle. Show that triangle ABC has a right angle at B.
Step 1: Find the gradient of side ABm_AB = (4 − 0) / (2 − (−1))m_AB = 4 / 3Step 2: Find the gradient of side BCm_BC = (1 − 4) / (6 − 2)m_BC = −3 / 4Step 3: Multiply the two gradientsm_AB × m_BC = (4/3) × (−3/4)= −12/12 = −1since m_AB × m_BC = −1, AB ⊥ BC, so the triangle has a right angle at B ✓to find the right angle in a triangle, check the gradients of the two sides meeting at each vertex — only one pair will multiply to −1
💡 Top tips
Always rearrange to y = mx + c first when comparing gradients. Don’t try to spot relationships in the messy ax + by + d = 0 form.
For perpendicular: flip the fraction, swap the sign. Quick and reliable. Check by multiplying — should give −1.
Once you have the right gradient, point-gradient form does the rest. Just plug in m and the point, then rearrange.
For perpendicular questions involving fractions, clear them early. Multiply by the denominator before moving things across.
Sketch the lines on a coordinate grid if you’re unsure. Parallel = same direction. Perpendicular = clear right-angle cross.
Watch for horizontal & vertical specials. A horizontal line (y = k) is perpendicular to any vertical line (x = h) — even though “m1·m2 = −1″ doesn’t apply.
Verify by plugging the given point into your final equation. If it doesn’t satisfy your equation, something went wrong.
⚠ Common mistakes
Forgetting to flip the fraction for perpendicular. Just changing the sign of m isn’t enough — you need to flip AND change sign. The negative of 2/3 is −2/3 (parallel-style), but the negative reciprocal is −3/2 (perpendicular).
Reading the gradient wrong from a non-standard form. Always rearrange to y = mx + c first.
Using the wrong line’s gradient. “Perpendicular to l1” means take l1‘s gradient and flip-and-negate it — don’t use a totally new gradient.
Treating the same-line case as parallel. If two equations turn out to be the same line (same gradient, same y-intercept), they’re not “parallel” in the usual sense — they’re identical.
Missing the negative sign. The product is −1, not +1. The minus is essential.
Forgetting that vertical lines have undefined gradient. Don’t try to write x = 5 in the form y = mx + c — it doesn’t fit.
Mixing up x and y subscripts when computing the gradient. (y2 − y1) on top, (x2 − x1) on the bottom — same order in both differences.
Parallel and perpendicular questions are some of the easiest marks in the IB exam — once you know the gradient rules, the rest is just plugging into point-gradient form. Get really fluent with the “flip the fraction, swap the sign” move because perpendicular gradients show up everywhere later: in normals to curves (calculus), perpendicular bisectors (geometry), and even vector dot products. The next section moves to quadratic functions, where the graph stops being a straight line and starts curving — but the same point-by-point thinking still applies.
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