Mathematics 0580 · Syllabus tracker
Co-Ordinate Geometry
Points, gradients, lengths, midpoints and the equation of a straight line, including parallel and perpendicular lines.
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Quick revision
Co-Ordinate Geometry on one sheet
For the night before. The full notes teach the topic; this is just what you must remember.
At a glance
Coordinates, gradient, length and midpoint, y = mx + c, and 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.
- x = 0: 2y = 12 → y = 6.
- 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
Rise 6, run 3: gradient 6 ÷ 3 = 2.
Worked example
Find the gradient of the line through (−1, 7) and (3, −5).
- Change in y: −5 − 7 = −12.
- Change in x: 3 − (−1) = 4.
- −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
Worked example
A is (−2, 1) and B is (4, 9). Find the midpoint and the length of AB.
- Midpoint: ((−2 + 4)/2, (1 + 9)/2) = (1, 5).
- 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
- Find m (from two points if needed).
- Substitute one point into y = mx + c.
- Solve for c.
- Write y = mx + c.
Worked example
Find the equation of the line through (2, 7) and (5, 13).
- m = (13 − 7) ÷ (5 − 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
Worked example
A(1, 2) and B(5, 4). Find the equation of the perpendicular bisector of AB.
- Midpoint (3, 3).
- Gradient of AB = 2/4 = 1/2, so perpendicular gradient −2.
- 3 = −2(3) + c → c = 9.
y = −2x + 9
✗Common mistakes
| Don’t write | Write 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.