Classifying a Quadrilateral from Coordinates
Mr. Delgado's students must classify quadrilateral ABCD with vertices A(1, 2), B(4, 6), C(8, 3), and D(5, −1). Tasha computes all four side lengths: AB = 5, BC = 5, CD = 5, and DA = 5. She then computes the diagonals: AC = √(7² + 1²) = √50 and BD = √(1² + (−7)²) = √50. Her conclusion reads: "All sides are equal, so it is a rhombus. The diagonals are also equal, which means it could be a rectangle, but a rectangle does not have four equal sides, so it is a rhombus and not a square." Mr. Delgado wants to correct the logic without simply saying "it's a square," because Tasha's numerical work was accurate and her error is in the implication, not the arithmetic.
1. Determine the correct classification using slopes, and state what Tasha failed to check.
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With all four sides equal, ABCD is a rhombus; the slopes settle whether it is also a square.
- Slope AB = $(6-2)/(4-1) = 4/3$
- Slope BC = $(3-6)/(8-4) = -3/4$
- Slope CD = $(-1-3)/(5-8) = 4/3$
- Slope DA = $(2-(-1))/(1-5) = -3/4$
Adjacent products: $(4/3)(-3/4) = -1$, so AB ⟂ BC, and the same holds at every vertex. All four sides are 5 and all four angles are 90°, so ABCD is a square.
Tasha never checked an angle. "All sides equal" gives a rhombus and nothing more; it leaves the angles free. She assumed equal diagonals were in tension with four equal sides, when in fact equal diagonals force the rhombus's angles toward 90°.
2. Explain why, in a rhombus, equal diagonals force the figure to be a square. Give a reasoning chain, not just a theorem name.
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Start from what a rhombus guarantees: the diagonals are perpendicular bisectors of each other, so they meet at a right angle and split into four right triangles with legs $p/2$ and $q/2$, where $p$ and $q$ are the diagonal lengths. Each side of the rhombus is the hypotenuse of such a triangle, so $s^2 = (p/2)^2 + (q/2)^2$.
If the diagonals are equal, $p = q$, then $s^2 = 2(p/2)^2 = p^2/2$, and each right triangle has two equal legs - it is isosceles. Each vertex angle of the rhombus is split into two 45° angles by a diagonal, so every vertex angle is 90°.
A rhombus with a right angle is a square, since adjacent angles are supplementary and all sides are already equal. So equal diagonals plus four equal sides is not a contradiction; the two facts together are a square. Tasha's sentence "it could be a rectangle, but a rectangle has unequal sides" is the mistaken step: a rectangle with equal sides is precisely a square.
3. Mr. Delgado wants a test procedure students can apply to any four coordinate points without guessing. Propose an efficient ordered checklist and justify its order.
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A good checklist stops as soon as a property fails, so cheap computations come first.
- Show it is a parallelogram. Compute the slopes of all four sides: AB and CD must match, and BC and DA must match. Equal opposite slopes mean parallel opposite sides. Alternatively, show one pair of opposite sides is both parallel and equal.
- Distinguish rhombus / rectangle. If consecutive slopes multiply to −1, there is a right angle, so the parallelogram is a rectangle; if not, it is only a parallelogram. Compute two adjacent side lengths: if the slopes were perpendicular and the adjacent sides are equal, it is a square; perpendicular with unequal sides gives a rectangle.
- If no right angle, compare adjacent side lengths. Equal gives a rhombus; unequal confirms a parallelogram.
Order matters because parallelism is a single slope comparison, while "square" is a compound claim needing both a right angle and equal sides. Doing slopes first also prevents Tasha's error: she checked lengths twice and never checked an angle, so the perpendicularity test was missing from her procedure entirely.