MISSION · Geometry
Coordinate Proofs
Put the shape on a grid and the proof becomes arithmetic
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The idea in one sentence
Place the figure on a coordinate grid with convenient letters for its corners, and every geometric claim turns into an algebraic one you can simply compute.
The three tools
Almost every coordinate proof is built from three formulas, and which one you reach for is decided by what you are trying to show.
| To prove | Use |
|---|---|
| segments are equal | the distance formula |
| lines are parallel | equal slopes |
| non-vertical lines are perpendicular | slopes multiplying to \(-1\) |
| a point bisects a segment | the midpoint formula |
Match the tool to the claim before you start computing. Working out all four for every pair of points produces a page of algebra and no argument.
Placement is the whole skill
You may put the figure anywhere, so put it where the algebra is easiest. Two rules:
Use the origin and the axes. A vertex at \((0,0)\) and a side along the \(x\)-axis kills a great many terms before they appear.
Use letters, not numbers, for the dimensions. A rectangle placed at \((0,0)\), \((a,0)\), \((a,b)\), \((0,b)\) proves something about every rectangle. Placed at \((0,0)\), \((6,0)\), \((6,4)\), \((0,4)\) it proves something about one rectangle, which is not a proof at all.
That second rule is where most marks are lost. Specific numbers are a check, not an argument.
A useful extra trick: use \(2a\) and \(2b\) rather than \(a\) and \(b\) when you expect to take midpoints. The halves cancel and you avoid fractions throughout.
A proof, start to finish
Claim. The diagonals of a rectangle are equal.
Placement. Vertices at \(A(0,0)\), \(B(a,0)\), \(C(a,b)\), \(D(0,b)\). This is a general rectangle: sides along the axes, and dimensions left as letters.
The computation. The diagonals are \(AC\) and \(BD\).
\(AC = \sqrt{(a-0)^2 + (b-0)^2} = \sqrt{a^2+b^2}\) \(BD = \sqrt{(0-a)^2 + (b-0)^2} = \sqrt{a^2+b^2}\)
The conclusion. The two are equal for every \(a\) and \(b\), so the diagonals of any rectangle are equal. \(\blacksquare\)
Notice how short that is next to the two-column version, which needs congruent triangles and several justified steps. Placement did the work.
Choosing between the two kinds of proof
Coordinate proofs are strong on distance, midpoint, parallel and perpendicular — the things formulas compute directly. The diagonals of a parallelogram bisecting each other, the midsegment of a triangle being half the third side, a triangle being right-angled: all quick.
They are weak on angles that are not right angles, and on circles, where the algebra grows faster than the insight.
A synthetic proof argues from congruence and similarity. It is better when the claim is about angle relationships, and it usually reveals why something is true rather than merely that it is.
Neither is more rigorous. They are different languages for the same statements, and Descartes’ insight was that the dictionary between them exists at all.
Common mix-ups
- Using specific numbers. It proves one case.
- Awkward placement. A vertex at \((3, 7)\) when it could have been at the origin turns a two-line proof into a long one.
- Over-specifying. Placing a general quadrilateral at \((0,0)\), \((a,0)\), \((a,b)\), \((0,b)\) has quietly assumed it is a rectangle, and then proves nothing about quadrilaterals.
- Forgetting the square root in the distance formula. For proving two lengths equal you can compare \(d^2\), which is legitimate and tidier — but say that you are doing it.
- Stopping at the algebra. Finish with a sentence stating what has been proved.
Check yourself
Which tool proves two segments equal, what do perpendicular slopes multiply to, and what is the midpoint of \((2,4)\) and \((8,10)\)?
Equal lengths call for the distance formula. For two non-vertical lines, perpendicular slopes multiply to \(-1\); a vertical and horizontal line are the separate exceptional pair. And the midpoint averages each coordinate: \(\left(\frac{2+8}{2}, \frac{4+10}{2}\right) = (5, 7)\).
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