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Cambridge IGCSE Mathematics — 0580 Extended

Topic 3.7: Coordinate geometry — Perpendicular Lines

Negative reciprocal gradients

Two lines are perpendicular if they meet at a right angle. Their gradients \(m_1\) and \(m_2\) satisfy

\[ m_1 \times m_2 = -1 \quad \Rightarrow \quad m_2 = -\dfrac{1}{m_1} \]

(provided neither line is vertical). A horizontal line (\(m = 0\)) is perpendicular to a vertical line (\(x = k\)).

The lines \(y = 2x\) and \(y = -\dfrac{1}{2}x\) are perpendicular. Why?

Perpendicular lines with gradients 2 and negative one half whose product is negative 1
\(2 \times \left(-\dfrac{1}{2}\right) = -1\).

Method

  1. Find \(m_1\) of the given line (rearrange if needed).
  2. The perpendicular gradient is \(m_2 = -\dfrac{1}{m_1}\).
  3. Substitute the given point into \(y - y_1 = m_2(x - x_1)\) and simplify.

Finding a perpendicular gradient

Always rearrange into \(y = mx + c\) first if the equation is given in another form.

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

Rearranging 2y equals 3x plus 1 to get m equals 3/2 then perpendicular gradient negative 2/3
\(m_1 = \dfrac{3}{2}\), so \(m_2 = -\dfrac{2}{3}\).

Paper 2 (non-calculator)

To find \(-\dfrac{1}{m}\) when \(m = \dfrac{a}{b}\), flip and change the sign: \(-\dfrac{b}{a}\). Check by multiplying: \(\dfrac{a}{b} \times \left(-\dfrac{b}{a}\right) = -1\).

Equation of a perpendicular through a point

Same idea as parallel lines, but use the negative reciprocal instead of the same gradient.

Find the equation of the line perpendicular to \(y = 2x + 1\) that passes through \((4, -1)\).

Perpendicular to y equals 2x plus 1 through (4, negative 1) giving y equals negative one half x plus 1
\(m_2 = -\dfrac{1}{2}\); \(y + 1 = -\dfrac{1}{2}(x - 4)\) simplifies to \(y = -\dfrac{1}{2}x + 1\).

Perpendicular bisector

The perpendicular bisector of the segment joining two points is the line that:

  • passes through the midpoint of the segment, and
  • has gradient equal to the negative reciprocal of the segment’s gradient.

Method

  1. Midpoint: \(M = \left(\dfrac{x_1 + x_2}{2},\, \dfrac{y_1 + y_2}{2}\right)\).
  2. Gradient of \(AB\): \(m_{AB} = \dfrac{y_2 - y_1}{x_2 - x_1}\).
  3. Perpendicular gradient: \(m = -\dfrac{1}{m_{AB}}\).
  4. Equation through \(M\) with that \(m\); simplify fully.

Find the equation of the perpendicular bisector of the line joining \(A(-3, 8)\) and \(B(9, -2)\).

Perpendicular bisector of A(negative 3, 8) and B(9, negative 2) through midpoint (3, 3) with gradient 6/5
\(M(3, 3)\); \(m_{AB} = -\dfrac{5}{6}\) so \(m = \dfrac{6}{5}\); equation \(y = \dfrac{6}{5}x - \dfrac{3}{5}\).

Try this

Find the gradient of a line perpendicular to \(y = -\dfrac{3}{4}x + 2\). Then find the equation of the line perpendicular to \(y = 3x - 1\) through \((0, 5)\).

Show answer
Answer

  1. Negative reciprocal of \(-\dfrac{3}{4}\)

    \[ m = -\dfrac{1}{-\tfrac{3}{4}} = \dfrac{4}{3} \]

  2. Perpendicular to \(y = 3x - 1\): \(m = -\dfrac{1}{3}\)

    \[ y - 5 = -\dfrac{1}{3}(x - 0) \]

  3. Fully simplified

    \[ y = -\dfrac{1}{3}x + 5 \]

Exam Traps

  • Do not confuse parallel (same \(m\)) with perpendicular (\(m_2 = -1/m_1\)).
  • Horizontal \(\perp\) vertical: a line \(y = k\) is perpendicular to \(x = c\), not to another horizontal line.
  • For a perpendicular bisector you need both the midpoint and the negative reciprocal — using only one loses marks.

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