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Mathematics 0580

Quick revision notes · Mathematics 0580 · Chapter 14 of 37

Co-Ordinate Geometry

Coordinates, gradient, length and midpoint, y = mx + c, and parallel and perpendicular lines.

5 sectionsWorked examples Common mistakes
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At a glance

Coordinates, gradient, length and midpoint, y = mx + c, and parallel and perpendicular lines.

  1. Coordinates and straight-line graphs
  2. Gradient
  3. Length and midpoint
  4. The equation y = mx + c
  5. Parallel and perpendicular lines

1Coordinates and straight-line graphs

  • (x, y): across first, then up. Points can be in any of the four quadrants.
  • Draw a line from its equation with a table of three x values (three points catch mistakes).
  • x = a is a vertical line; y = b is a horizontal line.
  • A line crosses the y-axis where x = 0 and the x-axis where y = 0.

Worked example

Find where 3x + 2y = 12 crosses the axes.

  1. x = 0: 2y = 12 → y = 6.
  2. y = 0: 3x = 12 → x = 4.

(0, 6) and (4, 0)

2Gradient

  • Gradient = change in y ÷ change in x (rise over run).
  • Uphill left to right → positive; downhill → negative; horizontal → 0; vertical → undefined.
  • From a graph: pick two points exactly on grid corners, far apart.

Formulas

m = (y₂ − y₁) / (x₂ − x₁)gradient between two points
xy(1, 2)(4, 8)run 3rise 6

Rise 6, run 3: gradient 6 ÷ 3 = 2.

Worked example

Find the gradient of the line through (−1, 7) and (3, −5).

  1. Change in y: −5 − 7 = −12.
  2. Change in x: 3 − (−1) = 4.
  3. −12 ÷ 4.

−3

3Length and midpoint

  • Midpoint: average the x's and average the y's.
  • Length: draw the right-angled triangle and use Pythagoras on the run and the rise.

Formulas

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)midpoint
d = √((x₂ − x₁)2 + (y₂ − y₁)2)length

Worked example

A is (−2, 1) and B is (4, 9). Find the midpoint and the length of AB.

  1. Midpoint: ((−2 + 4)/2, (1 + 9)/2) = (1, 5).
  2. Run 6, rise 8: √(36 + 64) = √100.

Midpoint (1, 5); length 10

4The equation y = mx + c

  • m is the gradient; c is where the line cuts the y-axis.
  • Rearrange to y = … before reading m: 2y = 6x + 4 → y = 3x + 2, so m = 3.
  • Through a point with gradient m: substitute the point into y = mx + c to find c (or use y − y₁ = m(x − x₁)).
  • Through two points: find m first, then c.

How to do it

  1. Find m (from two points if needed).
  2. Substitute one point into y = mx + c.
  3. Solve for c.
  4. Write y = mx + c.

Worked example

Find the equation of the line through (2, 7) and (5, 13).

  1. m = (13 − 7) ÷ (5 − 2) = 2.
  2. 7 = 2(2) + c → c = 3.

y = 2x + 3

5Parallel and perpendicular lines

  • Parallel lines have the same gradient.
  • Perpendicular gradients multiply to −1: flip the fraction and change the sign (2/3 → −3/2).
  • Perpendicular bisector of AB: passes through the midpoint of AB, with gradient perpendicular to AB.

Formulas

m₁ = m₂parallel
m₁ × m₂ = −1perpendicular

Worked example

A(1, 2) and B(5, 4). Find the equation of the perpendicular bisector of AB.

  1. Midpoint (3, 3).
  2. Gradient of AB = 2/4 = 1/2, so perpendicular gradient −2.
  3. 3 = −2(3) + c → c = 9.

y = −2x + 9

✗Common mistakes

Don’t writeWrite instead
Run over rise: (x₂ − x₁)/(y₂ − y₁) Always y on top: change in y ÷ change in x
Reading m from 2y = 6x + 4 as 6 Make y the subject first: y = 3x + 2, so m = 3
Perpendicular to m = 2 written as −2 Flip it too: −1/2
3 − (−2) = 1 3 − (−2) = 5. Put negatives in brackets

★In the exam

  • None of the gradient, midpoint or length formulas are given — learn them.
  • Sketch the points roughly first: it shows whether the gradient should be positive or negative.
  • Leave a length as a surd if the question asks for an exact answer.