Coordinate Proofs
Proving a theorem for every case by using variables instead of numbers.
The explanation
A coordinate proof puts a figure on the grid and proves something with formulas rather than with a two-column argument.
The critical move is placement. Put a vertex at the origin and a side along the x-axis. That kills as many variables as possible and makes the algebra manageable.
Use variables, not numbers. Proving something for the specific triangle (0,0), (4,0), (2,6) shows it holds for that one triangle. Using (0,0), (2a,0), (2b,2c) proves it for all of them.
Choose coefficients that keep midpoints clean. Writing 2a instead of a means the midpoint is a rather than a/2, which saves fractions throughout.
Then match the tool to the claim: distance for congruent, slope for parallel or perpendicular, midpoint for bisects.
A coordinate proof establishes a general result by assigning variable coordinates and verifying the claim algebraically. Generality requires that the placement impose no conditions beyond the figure's definition — placing a triangle at (0,0), (2a,0), (2b,2c) is fully general, since translation and rotation are rigid motions that preserve everything being proved, whereas placing it at (0,0), (2a,0), (a,2c) silently assumes isosceles.
Strategic placement uses the available rigid motions to zero out coordinates: one vertex at the origin and one side along an axis costs nothing and removes three variables. Doubling coefficients avoids fractions in midpoints.
The tool follows the claim: distance for congruence, slope for parallelism (equal) and perpendicularity (product −1), midpoint for bisection, and the midpoint of both diagonals for parallelogram tests.
Coordinate proofs are often shorter than synthetic ones — the triangle midsegment theorem takes three lines — but they are less illuminating about why a result holds, which is why both methods are taught.
Worked example
Prove the diagonals of a parallelogram bisect each other, using coordinates.
- Place it generally: A(0,0), B(a,0), C(a+b, c), D(b, c).
- Midpoint of AC: ((a+b)/2, c/2).
- Midpoint of BD: ((a+b)/2, c/2).
- The midpoints coincide.
Answer: Both diagonals share a midpoint, so each bisects the other.
Common mistakes
- Using specific numbers, which proves only one case.
- Placing the figure so that it accidentally assumes an extra property such as isosceles or right.