Classifying Figures Using Coordinates
Use slope, distance, and midpoint formulas to classify triangles and quadrilaterals placed on the coordinate plane. Prove that a figure is a parallelogram, rhombus, rectangle, or isosceles trapezoid using coordinate methods.
A lesson of Geometry, in Coordinate Geometry, about 10 minutes.
By the end of this lesson, you will be able to use slope, distance, and midpoint formulas to classify quadrilaterals and triangles on the coordinate plane, and verify geometric properties algebraically.
The Big Idea
Coordinate geometry converts geometric questions into algebra. Instead of constructing a proof with compass and straightedge, we compute slopes and distances to verify properties. This is both powerful — it works for any figure with known vertices — and elegant: every geometric property has an algebraic signature.
Tools for Coordinate Classification
| Property to verify | Tool |
|---|---|
| Sides are parallel | Slopes equal |
| Sides are perpendicular | Slopes are negative reciprocals |
| Sides are congruent | Distances equal |
| Diagonals bisect each other | Midpoints equal |
| Diagonals are congruent | Lengths equal |
| Diagonals are perpendicular | Slopes are negative reciprocals |
Classifying Quadrilaterals
Parallelogram: two pairs of parallel sides → equal slopes in both pairs.
Rectangle: parallelogram + perpendicular adjacent sides → slopes of adjacent sides are negative reciprocals.
Rhombus: parallelogram + all sides equal → all four side lengths equal.
Square: rectangle + rhombus.
Trapezoid: exactly one pair of parallel sides.
Isosceles trapezoid: trapezoid + equal legs.
Example 1 — Parallelogram
Classify quadrilateral with vertices $A(0,0)$, $B(5,0)$, $C(6,3)$, $D(1,3)$.
Slope $\overline{AB} = 0$; Slope $\overline{DC} = 0$ → $\overline{AB} \parallel \overline{DC}$
Slope $\overline{AD} = 3$; Slope $\overline{BC} = 3$ → $\overline{AD} \parallel \overline{BC}$
Two pairs parallel → parallelogram.
Adjacent slopes: $0$ and $3$ → product $= 0 \neq -1$ → not perpendicular → not a rectangle.
$AB = 5$, $AD = \sqrt{1 + 9} = \sqrt{10}$ → not equal → .