A coordinate grid turns geometry facts into algebra you can check. You have three tools: slope for parallel and perpendicular lines (from Parallel and Perpendicular Lines), the distance formula for lengths, and the midpoint formula. Together they can prove the facts from Rectangles, Rhombuses and Squares.
- Parallel sides: equal slopes. Perpendicular sides: slopes whose product is $-1$ (or one horizontal and one vertical).
- Congruent sides or diagonals: equal distances.
- Diagonals that bisect each other: the same midpoint.
Why it works
A proof with specific numbers shows one example. To prove a fact about every rectangle, place a general rectangle on the grid using letters: $(0, 0)$, $(a, 0)$, $(a, b)$, $(0, b)$. Any rectangle can be moved by a rigid motion to sit this way, and rigid motions keep lengths and angles, so the algebra covers them all. Putting a corner at the origin and sides on the axes makes most coordinates zero and the algebra short.
Worked example: Classify a quadrilateral
Classify $A(0, 0)$, $B(4, 3)$, $C(4, 8)$, $D(0, 5)$.
- Slopes: $AB$: $\frac{3}{4}$; $BC$: undefined (vertical); $CD$: $\frac{8 - 5}{4 - 0} = \frac{3}{4}$; $DA$: undefined. Both pairs of opposite sides are parallel: a parallelogram.
- Lengths: $AB = \sqrt{16 + 9} = 5$ and $BC = 5$. All four sides are 5: a rhombus.
- Right angle? $\overline{AB}$ (slope $\frac{3}{4}$) and $\overline{BC}$ (vertical) are not perpendicular.
Answer: A rhombus that is not a square.
Worked example: A general proof
Prove that the diagonals of any parallelogram bisect each other.
- Place it as $A(0, 0)$, $B(a, 0)$, $C(a + b, c)$, $D(b, c)$. Then $\overline{AB}$ and $\overline{DC}$ are both horizontal with length $a$, so opposite sides are parallel by design.
- Midpoint of $\overline{AC}$: $\left(\frac{a + b}{2}, \frac{c}{2}\right)$. Midpoint of $\overline{BD}$: $\left(\frac{a + b}{2}, \frac{c}{2}\right)$.
- The midpoints are the same point, so each diagonal cuts the other in half.
Answer: Proved for every parallelogram, because the letters stand for any values.
Worked example: Rectangle diagonals
Prove that the diagonals of a rectangle are congruent.
- Place the rectangle at $(0, 0)$, $(a, 0)$, $(a, b)$, $(0, b)$.
- One diagonal: from $(0, 0)$ to $(a, b)$, length $\sqrt{a^2 + b^2}$. The other: from $(a, 0)$ to $(0, b)$, length $\sqrt{a^2 + b^2}$.
Answer: The lengths are equal, so the diagonals are congruent.
Key idea
Use slopes for parallel and perpendicular, distances for congruent, midpoints for bisecting. For a general proof, place the figure at the origin along the axes and use letters for its coordinates.
Common mistake
Placing a figure at awkward coordinates, like $(3, 7)$, $(a, 2)$, $(b + 1, c)$. The algebra gets long and errors creep in, and specific numbers only prove one case. Put a vertex at the origin and a side on an axis, and use as few letters as the shape needs.