Geometry with an address system
Before Descartes, a geometrical fact had to be proved with a diagram and a chain of reasoning. After him, every point has an address (x, y) and geometrical questions become arithmetic. This chapter is the smallest useful piece of that idea.
The plane is divided by the two axes into four quadrants, numbered anticlockwise from the top right. The signs of the coordinates tell you which one you are in, and questions often ask for exactly that.
| Quadrant | x | y | Example |
|---|---|---|---|
| I (top right) | + | + | (3, 2) |
| II (top left) | − | + | (−3, 2) |
| III (bottom left) | − | − | (−3, −2) |
| IV (bottom right) | + | − | (3, −2) |
One triangle, three formulae
Take two points A(x₁, y₁) and B(x₂, y₂). Drop a horizontal line from A and a vertical line from B; they meet at a right angle. Every formula in this chapter is that one triangle read a different way.
The horizontal side has length x₂ − x₁. The vertical side has length y₂ − y₁. The hypotenuse is AB.
- |AB|
- the length of the segmentalways positive
- M
- the midpoint of ABaverage each coordinate separately
- m
- the gradient of ABundefined when x₂ = x₁
Move B so that it sits directly above A. The run becomes zero, so the gradient formula divides by zero — the line is vertical and has no gradient, which is not the same as a gradient of zero.
What the gradient tells you
The gradient is the single most useful number in the chapter, because it answers questions about parallel lines, perpendicular lines and collinear points without any drawing.
- Positive gradient — the line rises from left to right.
- Negative gradient — it falls from left to right.
- Zero gradient — horizontal. Undefined gradient — vertical. These two are opposites and are constantly confused.
- Parallel lines have equal gradients:
m₁ = m₂. - Perpendicular lines have gradients multiplying to −1:
m₁ m₂ = −1, so each is the negative reciprocal of the other.
Proving three points are collinear
Three points lie on one straight line if the gradient from the first to the second equals the gradient from the second to the third. Do not use the distance formula for this — checking that AB + BC = AC involves surds and is far more work for the same conclusion.
Worked example: a complete "show that" question
These questions look long but are entirely mechanical once you decide which formula answers which part. Write the coordinates down as (x₁, y₁) and (x₂, y₂) before substituting, and the sign errors disappear.
A(1, 2), B(5, 4) and C(3, 8) are three points. Show that triangle ABC is right-angled, and find the length of its hypotenuse.
- Gradient of AB:
(4 − 2)/(5 − 1) = 2/4 = 1/2.Take the points in a fixed order and keep it. Reversing both differences gives the same gradient, so consistency is all that matters. - Gradient of BC:
(8 − 4)/(3 − 5) = 4/(−2) = −2. (1/2) × (−2) = −1, so AB is perpendicular to BC and the right angle is at B.The product being −1 is the test. Saying which vertex holds the right angle is usually worth a mark.- The hypotenuse is the side opposite the right angle, namely AC.The right angle is at B, so the hypotenuse must be the side not touching B.
|AC| = √[(3 − 1)² + (8 − 2)²] = √(4 + 36) = √40 = 2√10.Simplify the surd — √40 alone will usually cost the last mark.
Right-angled at B; hypotenuse AC = 2√10 ≈ 6.32 units
The equation of a straight line
Once you have a gradient and a point, you have the line. The two forms below are the ones the paper expects, and converting between them is a matter of expanding a bracket.
The slope-intercept form y = mx + c is the one to use when you are told the y-intercept. The point-slope form y − y₁ = m(x − x₁) is the one to use when you are told a point, which is far more common.
Before you leave this chapter
- Distance is Pythagoras; midpoint is two averages; gradient is rise over run. All three come from the same triangle.
- Horizontal → gradient 0. Vertical → gradient undefined. Never write "gradient = ∞" in an answer.
- Parallel ⟺ equal gradients. Perpendicular ⟺ product of gradients is −1.
- Collinear points are proved with gradients, not distances.
- Given a point and a gradient, use y − y₁ = m(x − x₁) and expand.
Naming a shape from its coordinates
A standard long question gives you four points and asks what kind of quadrilateral they form. There is a fixed order of checks, and following it stops you from claiming more than you have proved.
Work out the four gradients first, then any lengths you need. Each property you establish narrows the answer, and the final name is whichever description you can fully justify — never guess from the look of your sketch.
| To prove it is a… | You must show | Tool |
|---|---|---|
| Trapezium | exactly one pair of opposite sides parallel | gradients |
| Parallelogram | both pairs of opposite sides parallel | gradients |
| Rhombus | a parallelogram with all four sides equal | gradients, then distance |
| Rectangle | a parallelogram with one right angle | gradients, then m₁m₂ = −1 |
| Square | all four sides equal and one right angle | distance and m₁m₂ = −1 |
Do not claim a rectangle without checking the angle
Equal-and-parallel opposite sides prove a parallelogram and nothing more. A rhombus also has equal sides but no right angle, and a rectangle has a right angle but not necessarily equal sides. Each extra word in the name costs one extra check, and the mark scheme allots a mark to each.