Home / IGCSE Maths 0580 / Coordinate geometry / Perpendicular lines and bisectors
Cambridge IGCSE · Topic 3.7
IGCSE Maths 0580 / Topic 3 / IGCSE

Flip it and change the sign.

Extended. Lines that meet at a right angle have gradients that multiply to \(-1\). That one fact gives you perpendicular lines through any point and the perpendicular bisector of any segment.

Full-screen labStraight lines lab → Topic 3All coordinate geometry lessons →
The key idea

If two lines are perpendicular, \(m_1 \times m_2 = -1\), so the perpendicular gradient is \(-\dfrac{1}{m}\): flip the fraction and change the sign. The perpendicular bisector of \(AB\) passes through the midpoint of \(AB\) with the perpendicular gradient.

−1 1 2 3 4 5 6 −3 −2 −1 1 2 3 4 5 6 7 x y A(0, 6) B(4, −2) M(2, 2) AB: gradient −2 bisector: gradient 1/2 (−2 × 1/2 = −1) y = x/2 + 1
Fig. 1: AB has gradient -2; its perpendicular bisector passes through the midpoint M(2, 2) with gradient 1/2, giving y = x/2 + 1
Section 01

Gradients that multiply to −1.

In the lab, rotate the first line and watch the perpendicular line turn with it, always at \(90^\circ\). Read both gradients each time and multiply them. Try a steep line, a shallow one and a downhill one: the product never changes.

Worked example Perpendicular gradients
Find the gradient of a line perpendicular to (a) \(y = 3x - 2\) (b) \(2y = 3x + 1\).
  1. (a) The gradient is 3. Flip: \(\dfrac{1}{3}\). Change sign: \(-\dfrac{1}{3}\).
  2. (b) Make \(y\) the subject: \(y = \dfrac{3}{2}x + \dfrac{1}{2}\), so the gradient is \(\dfrac{3}{2}\).
  3. Flip and change sign: \(-\dfrac{2}{3}\). Check: \(\dfrac{3}{2} \times \left(-\dfrac{2}{3}\right) = -1\).
Answer: (a) \(-\dfrac{1}{3}\)   (b) \(-\dfrac{2}{3}\)
Worked example Testing for a right angle
Show that the triangle with vertices \(P(1, 2)\), \(Q(4, 3)\) and \(R(2, 9)\) has a right angle at \(Q\).
  1. Gradient \(PQ = \dfrac{3 - 2}{4 - 1} = \dfrac{1}{3}\).
  2. Gradient \(QR = \dfrac{9 - 3}{2 - 4} = \dfrac{6}{-2} = -3\).
  3. \(\dfrac{1}{3} \times (-3) = -1\), so \(PQ\) is perpendicular to \(QR\).
Answer: Angle \(PQR = 90^\circ\)
Examiner trap

Doing only half the recipe. The perpendicular to gradient 3 is not \(-3\) (sign only) and not \(\dfrac{1}{3}\) (flip only). It is \(-\dfrac{1}{3}\). Multiply your two gradients: you must get \(-1\).

Section 02

Perpendicular through a point.

Find the perpendicular gradient first, then substitute the given point into \(y = mx + c\) to find \(c\), exactly as for parallel lines.

Worked example From y = mx + c
Find the equation of the line perpendicular to \(y = 3x - 2\) that passes through \((6, 1)\).
  1. Perpendicular gradient: \(-\dfrac{1}{3}\).
  2. \(1 = -\dfrac{1}{3} \times 6 + c = -2 + c\), so \(c = 3\).
  3. Multiply \(y = -\dfrac{1}{3}x + 3\) by 3 and rearrange for a form without fractions: \(x + 3y = 9\).
Answer: \(y = -\dfrac{1}{3}x + 3\), or \(x + 3y = 9\)
Worked example From ax + by = c
Find the equation of the line perpendicular to \(4x - 5y = 10\) that passes through \((4, -3)\).
  1. \(-5y = -4x + 10\), so \(y = \dfrac{4}{5}x - 2\). The gradient is \(\dfrac{4}{5}\).
  2. Perpendicular gradient: \(-\dfrac{5}{4}\).
  3. \(-3 = -\dfrac{5}{4} \times 4 + c = -5 + c\), so \(c = 2\).
  4. \(y = -\dfrac{5}{4}x + 2\); multiplying by 4: \(5x + 4y = 8\). Check: \(20 - 12 = 8\).
Answer: \(y = -\dfrac{5}{4}x + 2\), or \(5x + 4y = 8\)
Your turn
Find the equation of the line perpendicular to \(y = -2x + 7\) that passes through \((4, 5)\).
Examiner trap

Using the original gradient by mistake when you substitute. Write the NEW gradient down on its own line before finding \(c\), so you cannot slip back to the old one.

Stage 1 · Learn

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A line has gradient 4. What is the gradient of a line perpendicular to it?

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What is the gradient of a line perpendicular to a line with gradient \(-\dfrac{2}{3}\)?

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Which pair of lines is perpendicular?

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The perpendicular bisector of \(AB\) always passes through

Section 03

The perpendicular bisector.

The perpendicular bisector of \(AB\) cuts \(AB\) in half at right angles. Every point on it is the same distance from \(A\) as from \(B\). It needs two facts: it passes through the midpoint, and its gradient is perpendicular to \(AB\).

Worked example A bisector through a whole-number midpoint
Find the equation of the perpendicular bisector of \(AB\), where \(A(0, 6)\) and \(B(4, -2)\).
  1. Midpoint \(M = \left(\dfrac{0 + 4}{2}, \dfrac{6 + (-2)}{2}\right) = (2, 2)\).
  2. Gradient \(AB = \dfrac{-2 - 6}{4 - 0} = -2\), so the perpendicular gradient is \(\dfrac{1}{2}\).
  3. \(2 = \dfrac{1}{2} \times 2 + c = 1 + c\), so \(c = 1\).
Answer: \(y = \dfrac{1}{2}x + 1\) (Fig. 1)
Worked example Fractions in the answer
Find the equation of the perpendicular bisector of the line joining \((-3, 8)\) and \((9, -2)\).
  1. Midpoint: \(\left(\dfrac{-3 + 9}{2}, \dfrac{8 + (-2)}{2}\right) = (3, 3)\).
  2. Gradient of the line: \(\dfrac{-2 - 8}{9 - (-3)} = \dfrac{-10}{12} = -\dfrac{5}{6}\). Perpendicular gradient: \(\dfrac{6}{5}\).
  3. \(3 = \dfrac{6}{5} \times 3 + c = \dfrac{18}{5} + c\), so \(c = -\dfrac{3}{5}\).
  4. \(y = \dfrac{6}{5}x - \dfrac{3}{5}\); multiplying by 5: \(5y = 6x - 3\).
Answer: \(y = \dfrac{6}{5}x - \dfrac{3}{5}\), or \(6x - 5y = 3\)
Examiner trap

Putting \(A\) or \(B\) into the equation instead of the midpoint. The bisector passes through \(M\), not through the end points, so substitute the midpoint to find \(c\).

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Find the gradient of a line perpendicular to \(2y = 3x + 1\).

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Find the equation of the line perpendicular to \(y = 3x - 2\) that passes through \((6, 1)\).

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Find the equation of the perpendicular bisector of \(AB\), where \(A(-2, 1)\) and \(B(6, 5)\).

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The lines \(3x + 2y = 7\) and \(y = kx + 4\) are perpendicular. Find \(k\).

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Six original Cambridge-style questions
Perpendicular gradients, proving lines meet at \(90^\circ\), perpendicular lines through a point, a perpendicular bisector and the shortest path from a house to a road.