7.4 Coordinate Proofs, Perimeter & Area

Proving Figures with Coordinates

Standards GEO-G.GPE.4 (prove simple geometric theorems algebraically) and GEO-G.GPE.7 (compute perimeters and areas from coordinates) power the coordinate-proof questions that often appear in Parts II-IV. A full-credit coordinate proof states which property you are testing, shows the calculation, and writes a conclusion — the calculation alone will not earn full credit.

Three tools do all the work:

• Slope proves sides parallel (equal slopes) or perpendicular / right angle (slopes multiply to −1).
• Distance proves sides congruent (equal lengths).
• Midpoint proves diagonals bisect each other (shared midpoint).

What to test for each figure

Worked Coordinate Proof: ABCD Is a Square

Prove that quadrilateral ABCD with A(−2, 1), B(4, 3), C(6, −3), D(0, −5) is a square.

Step 1 — right angle (slope). Slope of AB = (3 − 1)/(4 − (−2)) = 2/6 = 1/3. Slope of BC = (−3 − 3)/(6 − 4) = −6/2 = −3. Product = (1/3)(−3) = −1, so AB ⊥ BC and angle B is a right angle.

Step 2 — four congruent sides (distance). AB = √(6² + 2²) = √40. BC = √(2² + 6²) = √40. CD = √(6² + 2²) = √40. DA = √(2² + 6²) = √40. All four sides equal √40.

Step 3 — conclusion. ABCD has four congruent sides (making it a rhombus) and a right angle (making it a rectangle). A figure that is both a rhombus and a rectangle is a square. ∎

Perimeter and Area on the Grid

Perimeter is the sum of the side lengths from the distance formula. For the square above, perimeter = 4·√40 = 4·(2√10) = 8√10.

Area can be found three ways, and the Regents accepts any correct method:

• Base × height when a side is horizontal or vertical. A right triangle with legs 8 and 6 has area = ½·8·6 = 24.
• Decomposition / bounding box: enclose the figure in a rectangle and subtract the surrounding right triangles.
• Shoelace formula: list the vertices in order (clockwise or counterclockwise) and compute Area = ½ |Σ(xᵢ·yᵢ₊₁ − xᵢ₊₁·yᵢ)|.

Shoelace worked example: triangle A(0, 0), B(6, 0), C(0, 9). Area = ½|(0·0 − 6·0) + (6·9 − 0·0) + (0·0 − 0·9)| = ½|0 + 54 − 0| = 27. The base 6 and height 9 confirm ½·6·9 = 27.

Area by decomposition (bounding box): When no side is horizontal or vertical, enclose the figure in a rectangle and subtract the corner triangles. For triangle P(1, 1), Q(5, 3), R(2, 6), the bounding box runs x = 1 to 5 and y = 1 to 6 (area 4 × 5 = 20). Subtract the three right triangles: ½·4·2 = 4, ½·3·3 = 4.5, and ½·1·5 = 2.5. The area is 20 − 4 − 4.5 − 2.5 = 9, matching the shoelace result, so use whichever is faster under time pressure.

Proving a Triangle Is Right or Isosceles

Triangles use the same three tools. For A(−4, 2), B(8, 6), C(2, −6): CA = √(6² + 8²) = √100 = 10, AB = √(12² + 4²) = √160, and BC = √(6² + 12²) = √180. No two sides match, so the triangle is scalene. Testing angle A, slope CA = 8/(−6) = −4/3 and slope AB = 4/12 = 1/3, whose product is −4/9 (not −1), so there is no right angle at A. To prove isosceles, you would instead show two distances equal; to prove right, you would show two slopes multiply to −1.

Simplifying Radical Answers

Regents answer choices almost always show radicals in simplest form, so simplify last. A perimeter of 4√40 becomes 4·2√10 = 8√10, and a side of √180 becomes 6√5. Leaving √40 or √180 unsimplified is a frequent way to miss the correct choice even after the geometry is right.

Diagram: Classifying a Quadrilateral

Use the flow below to decide the most specific name once you have the coordinates.

Common traps

• Proving a figure is a parallelogram and stopping — that is not enough for rectangle, rhombus, or square.
• Skipping the written justification; Parts II-IV award credit for naming the property, not just the arithmetic.
• Forgetting the absolute value in the shoelace formula, or listing vertices out of order.
• Confusing "most specific" classification: four equal sides alone gives a rhombus, not a square, unless a right angle is also shown.