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
- Flip the fraction (reciprocal).
- Change the sign.
- 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
- Parallel ⇒ same gradient.
- Write y = 2x + c and substitute (1, 4).
- 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
- Perpendicular ⇒ negative reciprocal of 2.
- Write y = −½x + c and substitute (4, 1).
- 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
- Gradient formula for [QR] — the two points L is NOT through.
- L is perpendicular to QR, so take the negative reciprocal (flip and change the sign).
- 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
- 'Passes through (k, 3)' means x = k and y = 3 fit the equation of L.
- 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 = …').