Coordinate Proofs
Place geometric figures strategically on the coordinate plane to simplify proofs (e.g., one vertex at origin, one side along the x-axis). Write complete coordinate proofs for theorems about quadrilaterals and triangles.
A lesson of Geometry, in Coordinate Geometry, about 10 minutes.
By the end of this lesson, you will be able to write coordinate proofs by strategically placing figures on the coordinate plane, using the distance, midpoint, and slope formulas as algebraic justification.
The Big Idea
A coordinate proof positions a geometric figure on the coordinate plane using carefully chosen coordinates, then uses algebraic formulas (distance, midpoint, slope) to prove statements about the figure. The power of coordinate proofs is that they convert geometry into algebra — valid for all figures of the given type, not just one specific case.
Strategic placement is the art: choosing coordinates that minimize computation while preserving generality.
Placement Strategies
Place one vertex at the origin — eliminates constants from that vertex's coordinates.
Align sides with the axes — horizontal/vertical sides have slope 0 or undefined, simplifying slope calculations.
Use even numbers for lengths — if the midpoint formula will be used, using even coordinates ensures integer midpoints.
Keep it general — use variables ($a$, $b$, $c$) rather than specific numbers to prove statements for all cases.
Strategic Coordinate Placements
| Figure | Good placement |
|---|---|
| Right triangle | Right angle at origin; legs along axes |
| Rectangle | One vertex at origin; sides along axes |
| Parallelogram | One vertex at origin; use $(a, 0)$, $(b, c)$, $(a+b, c)$ |
| Isosceles triangle | Base on $x$-axis, symmetric about $y$-axis: $(-a, 0)$, $(a, 0)$, $(0, b)$ |
Example 1 — Midpoint of the Hypotenuse (Right Triangle)
Prove: the midpoint of the hypotenuse of a right triangle is equidistant from all three vertices.
Setup: Place the right angle at the origin, legs along the axes.
$A(0, 0)$, $B(2a, 0)$, $C(0, 2b)$
Find midpoint $M$ of hypotenuse $\overline{BC}$:
$$M = \left(\frac{2a + 0}{2}, \frac{0 + 2b}{2}\right) = (a, b)$$
Compute distances:
$MA = \sqrt{a^2 + b^2}$
$MB = \sqrt{(2a-a)^2 + (0-b)^2} = \sqrt{a^2 + b^2}$
$MC = \sqrt{(0-a)^2 + (2b-b)^2} = \sqrt{a^2 + b^2}$
$MA = MB = MC$ → midpoint of hypotenuse equidistant from all three vertices ∎