Classifying a Quadrilateral from Coordinates
A student plots quadrilateral ABCD with vertices A(-3, 1), B(1, 4), C(5, 1), and D(1, -2). She computes the slopes of the diagonals AC and BD. AC is horizontal, and BD is vertical, so the diagonals are perpendicular. She also notices that the diagonals bisect each other at (1,1). She concludes that ABCD must be a square because 'perpendicular diagonals that bisect each other make a square.' Her partner says the figure is a rhombus, not necessarily a square, and points out that the sides are all length 5. The student replies that a rhombus with perpendicular diagonals is always a square. The teacher asks the pair to settle the disagreement with coordinate evidence: side lengths, slopes of adjacent sides, and the definition of a square. The student is allowed to use the distance formula and slope formula, but not a graph. She must also explain whether the perpendicular-diagonal fact is enough by itself.
1. Use the distance formula to find AB, BC, CD, and DA. What does this prove about the quadrilateral, and what does it not prove?
Read the worked answer
AB = sqrt((1-(-3))^2 + (4-1)^2) = sqrt(4^2+3^2)=sqrt(16+9)=sqrt25=5. BC = sqrt((5-1)^2+(1-4)^2)=sqrt(4^2+(-3)^2)=5. CD = sqrt((1-5)^2+(-2-1)^2)=sqrt((-4)^2+(-3)^2)=5. DA = sqrt((-3-1)^2+(1-(-2))^2)=sqrt((-4)^2+3^2)=5.
All four sides are congruent, so ABCD is a rhombus. This proves the figure is equilateral, but it does not prove it is a square. A square needs four right angles in addition to four congruent sides. The side-length calculation alone cannot rule out a non-square rhombus, because a rhombus can have equal sides without right angles. To test for a square, we need to check an angle, for example by comparing slopes of adjacent sides.
2. Use slopes to determine whether any adjacent sides are perpendicular. Then classify ABCD as specifically as the evidence allows. Show calculations.
Read the worked answer
Slope AB = (4-1)/(1-(-3)) = 3/4. Slope BC = (1-4)/(5-1) = -3/4. Slope CD = (-2-1)/(1-5) = -3/-4 = 3/4. Slope DA = (1-(-2))/(-3-1) = 3/-4 = -3/4.
Adjacent slopes: (3/4)(-3/4) = -9/16, not -1. So AB is not perpendicular to BC. The same product occurs at each vertex, so there are no right angles. The figure has four congruent sides and no right angles, so the most specific classification is a rhombus that is not a square. The diagonals are perpendicular (AC horizontal, BD vertical), but that is a property of a rhombus, not enough to make it a square. The student's partner is correct: it is a rhombus, not necessarily a square, and in this case definitely not a square.
3. Evaluate the student's claim that 'a rhombus with perpendicular diagonals is always a square.' Is the claim true or false? Use the coordinates and a general property of rhombi to explain. What single additional check would settle whether ABCD is a square?
Read the worked answer
The claim is false. In a rhombus, the diagonals are always perpendicular, so adding the condition 'perpendicular diagonals' does not distinguish a square from a non-square rhombus. For ABCD, AC is horizontal from (-3,1) to (5,1) with length 8, and BD is vertical from (1,4) to (1,-2) with length 6. The diagonals are perpendicular, but they are not congruent. A square is a rhombus with congruent diagonals (or, equivalently, with right angles).
The single additional check that settles it is to compare the diagonals: AC = 8 and BD = 6, so they are not equal. Therefore ABCD is not a square. Checking one angle would also work: the product of adjacent slopes is -9/16, not -1, so no right angle exists. The student has confused a necessary condition of a square with a sufficient one.