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3 · E3.1–E3.7

Coordinate geometry

Topic 3 · Core and Extended · Papers 1–4

🎯What you need to be able to do

  • Plot and read coordinates, and draw straight-line graphs such as \( y = 7 - 4x \) and \( 3x + 2y = 5 \).
  • Find the gradient of a line, from a graph or from two points.
  • Find and interpret the equation of a line in the forms \( y = mx + c \), \( ax + by = c \) and \( x = k \).
  • Find the equation of a line parallel to a given line.
  • Find the length and midpoint of a line segment EXTENDED.
  • Find the equation of a line perpendicular to a given line, including a perpendicular bisector EXTENDED.

📚The mathematics

Gradient and \( y = mx + c \)

gradient \( m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} = \dfrac{\text{rise}}{\text{run}} \)
\( y = mx + c \): gradient \(m\), \(y\)-intercept \(c\)
\( x = k \): vertical line; \( y = k \): horizontal line

Uphill left to right is a positive gradient, downhill negative. To read \(m\) and \(c\) from an equation such as \( 3x + 2y = 12 \), make \(y\) the subject first: \( y = -1.5x + 6 \). To find the equation through a point with a known gradient, substitute the point into \( y = mx + c \) to find \(c\). The syllabus expects equations in a fully simplified form.

To draw a line, work out three points (three, so a mistake shows up as a point off the line), or use the intercepts.

A straight line crossing the y-axis at 3 and passing through (2, minus 1). A gradient triangle shows a run of 2 and a fall of 4, giving gradient minus 2, so the equation is y = minus 2x + 3.
Practice question 4: intercept 3, gradient \( -4 \div 2 = -2 \), so \( y = -2x + 3 \).

Parallel and perpendicular lines

Parallel lines have equal gradients. EXTENDED Perpendicular lines have gradients that multiply to \(-1\): the perpendicular gradient is the negative reciprocal, so \( \tfrac{4}{3} \) becomes \( -\tfrac{3}{4} \) and \(-2\) becomes \( \tfrac{1}{2} \).

Length and midpoint EXTENDED

midpoint \( \left(\dfrac{x_{1} + x_{2}}{2}, \dfrac{y_{1} + y_{2}}{2}\right) \)
length \( \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \) (Pythagoras)

A perpendicular bisector of \(AB\) goes through the midpoint of \(AB\) with the perpendicular gradient.

Points A(minus 2, 1) and B(4, 9) joined by a line segment with a gradient triangle of run 6 and rise 8. The midpoint M(1, 5) is marked, and the perpendicular bisector 3x + 4y = 23 crosses AB at M at right angles.
The worked example: gradient \( \tfrac{8}{6} = \tfrac{4}{3} \), length 10, midpoint \( (1, 5) \), and the perpendicular bisector \( 3x + 4y = 23 \).

✏️Worked example (non-calculator)

\(A\) is \( (-2, 1) \) and \(B\) is \( (4, 9) \). (a) Find the gradient of \(AB\) and the equation of the line \(AB\). [3] (b) Find the equation of the line parallel to \(AB\) through \( (0, -2) \). [1] (c) EXTENDED Find the length of \(AB\) and the equation of its perpendicular bisector. [5]

(a) \( m = \dfrac{9 - 1}{4 - (-2)} = \dfrac{8}{6} = \dfrac{4}{3} \). Substituting \( (4, 9) \) into \( y = \tfrac{4}{3}x + c \): \( 9 = \tfrac{16}{3} + c \), so \( c = \tfrac{11}{3} \). The line is \( y = \tfrac{4}{3}x + \tfrac{11}{3} \), or \( 4x - 3y = -11 \).

(b) Same gradient, intercept \(-2\): \( y = \tfrac{4}{3}x - 2 \).

(c) \( AB = \sqrt{6^{2} + 8^{2}} = 10 \). Midpoint \( (1, 5) \); perpendicular gradient \( -\tfrac{3}{4} \). \( y - 5 = -\tfrac{3}{4}(x - 1) \Rightarrow 4y - 20 = -3x + 3 \Rightarrow 3x + 4y = 23 \).

Check it. Both points should satisfy (a): \( 4(-2) - 3(1) = -11 \) ✓ and \( 16 - 27 = -11 \) ✓. The midpoint should satisfy (c): \( 3 + 20 = 23 \) ✓.
Run over rise. Putting the \(x\)-difference on top gives \( \tfrac{6}{8} \), which happens to be the perpendicular gradient’s size — and so looks plausible. Always \(y\) over \(x\).

📝Practise

Questions in the style of the current papers. EXTENDED marks Extended-only content.

1. (Non-calculator.) Find the gradient of the line through \( (1, 7) \) and \( (5, -1) \). [2]
\( \dfrac{-1 - 7}{5 - 1} = \dfrac{-8}{4} = -2 \).
2. (Non-calculator.) Find the gradient and the \(y\)-intercept of the line \( 3x + 2y = 12 \). [2]
\( y = -1.5x + 6 \): gradient \(-1.5\), \(y\)-intercept 6.
3. (Non-calculator.) Find the equation of the line parallel to \( y = 3x - 2 \) that passes through \( (2, 1) \). [2]
\( y = 3x + c \); \( 1 = 6 + c \Rightarrow c = -5 \). \( y = 3x - 5 \).
4. (Non-calculator.) A straight line crosses the \(y\)-axis at \( (0, 3) \) and passes through \( (2, -1) \). Find its equation. [2]
Gradient \( \dfrac{-1 - 3}{2 - 0} = -2 \), intercept 3: \( y = -2x + 3 \). (See the first diagram.)
5. (Non-calculator.) Does the point \( (4, 10) \) lie on the line \( y = 2x + 3 \)? Show how you decide. [1]
When \( x = 4 \), \( y = 11 \), not 10, so it does not.
6. (Non-calculator.) EXTENDED \(P\) is \( (-3, 4) \) and \(Q\) is \( (5, -2) \). Find the length of \(PQ\) and the coordinates of its midpoint. [3]
\( \sqrt{8^{2} + 6^{2}} = 10 \); midpoint \( (1, 1) \).
7. (Non-calculator.) EXTENDED Find the gradient of a line perpendicular to \( 2y = 5x + 1 \). [2]
\( y = 2.5x + 0.5 \), gradient \( \tfrac{5}{2} \); perpendicular gradient \( -\tfrac{2}{5} \).
8. (Non-calculator.) EXTENDED Find the equation of the perpendicular bisector of the line joining \( (1, 7) \) and \( (7, -1) \). [4]
Midpoint \( (4, 3) \); gradient \( \tfrac{-8}{6} = -\tfrac{4}{3} \), so perpendicular gradient \( \tfrac{3}{4} \). \( y - 3 = \tfrac{3}{4}(x - 4) \Rightarrow y = \tfrac{3}{4}x \) — it happens to pass through the origin.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Desmos — plot a line and its perpendicular to see the right angle (use equal axis scales)
  • Corbettmaths — equations of straight lines, parallel and perpendicular