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097e145
NotesMath AATopic 2.1Perpendicular lines
Back to Math AA Topics
2.1.32 min read

Perpendicular lines

IB Mathematics: Analysis and Approaches • Unit 2

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Contents

  • Perpendicular lines — negative reciprocal
  • Build a line ∥ or ⊥ through a point
  • Putting it together — a full exam question
Perpendicular ⇒ m₁ × m₂ = −1: Two lines are perpendicular (they meet at 90°) when their gradients multiply to −1.

So the second gradient is m₂ = −1/m₁ — flip the fraction and change the sign (the negative reciprocal).

Flip between the chips: parallel lines keep the same gradient; for perpendicular, the second gradient is the negative reciprocal, turning the line exactly 90°.

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IB-style question — the perpendicular gradient

A line has gradient ⅔.

Find the gradient of a line perpendicular to it.

Step by step

  1. Flip the fraction (reciprocal).
  2. Change the sign.
  3. Check the product is −1.

Final answer

Perpendicular gradient = −3/2.

Whole numbers too: Treat a whole number as a fraction over 1: a gradient of 5 = 5/1, so the perpendicular gradient is −1/5.

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Get the gradient, then substitute the point: Same recipe both times: find the required gradient — the same one for parallel, the negative reciprocal for perpendicular — then put it into y = mx + c and substitute the point to find c.

IB-style question — parallel through a point

Find the equation of the line through (1, 4) parallel to y = 2x + 7.

Step by step

  1. Parallel ⇒ same gradient.
  2. Write y = 2x + c and substitute (1, 4).
  3. Solve for c.

Final answer

y = 2x + 2.

IB-style question — perpendicular through a point

Find the equation of the line through (4, 1) perpendicular to y = 2x − 3.

Step by step

  1. Perpendicular ⇒ negative reciprocal of 2.
  2. Write y = −½x + c and substitute (4, 1).
  3. Solve for c.

Final answer

y = −½x + 3.

Faster alternative — point–gradient form: Another standard method (also in the formula booklet) is point–gradient form y − y₁ = m(x − x₁): drop in the point and gradient and expand.

Same answer — use whichever you prefer.

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Spot this shape — it appears most years: Three points, one perpendicular line, one unknown — always the same three moves, in order:

You're given

  • 3 points (a triangle)
  • and a side, e.g. [QR]

You build

  • the line through one point
  • perpendicular to that side

You use it

  • put a point into the line
  • → solve for the unknown k

Step through it with the buttons: plot the points → gradient of QR → flip it for L → draw L through P → read off k.

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Part (a) — find the equation of L

The points are P(1, 9), Q(2, 1) and R(6, 3).

The line L passes through P and is perpendicular to [QR].

Find the equation of L.

Step by step

  1. Gradient formula for [QR] — the two points L is NOT through.
  2. L is perpendicular to QR, so take the negative reciprocal (flip and change the sign).
  3. Build L through P(1, 9) with point–gradient form, then tidy.

Final answer

y = −2x + 11.

Part (b) — L passes through (k, 3), find k

Using the line L: y = −2x + 11 from part (a), L also passes through the point (k, 3).

Find the value of k.

Step by step

  1. 'Passes through (k, 3)' means x = k and y = 3 fit the equation of L.
  2. Solve for k.

Final answer

k = 4.

See how every part is linked: gradient of QR → perpendicular gradient → equation of L → use the equation. Each arrow adds just one small rule, and part (b) isn't a new idea — it's 'put the point into the line you already built'.

Keep the steps in order and you can't get lost.
Three things that catch people out: 1. Take the gradient of the segment (the two points the line is NOT through) — not of one of them with P.

2. Perpendicular = flip and change the sign (½ → −2, not −½).

3. For '(k, 3)', substitute x = k, y = 3 — and write the answer as y = … (the scheme rejects 'L = …').

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Find the gradient of a line perpendicular to y = (3/4)x + 1. [2 marks]

Related Math AA Topics

Continue learning with these related topics from the same unit:

2.1.1Equations of lines
2.1.2Parallel lines
2.1.4Perpendicular bisector
2.2.1Function notation
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