Question

Determine the most precise name for ABCD (parallelogram, rhombus, rectangle, or square). Explain how you determined your answer. You must support your answer using length or slope.
A(3, 5), B(7, 6), C(6, 2), D(2, 1)

Answer 1

Answer:  Rhombus

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Reason:

Let's find the distance from A to B. This is equivalent to finding the length of segment AB. I'll use the distance formula.

$A = (x_1,y_1) = (3,5) \text{ and } B = (x_2, y_2) = (7,6)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-7)^2 + (5-6)^2}\\\\d = \sqrt{(-4)^2 + (-1)^2}\\\\d = \sqrt{16 + 1}\\\\d = \sqrt{17}\\\\d \approx 4.1231$

Segment AB is exactly $\sqrt{17}$ units long, which is approximately 4.1231 units.

If you were to repeat similar steps for the other sides (BC, CD and AD) you should find that all four sides are the same length. Because of this fact, we have a rhombus.

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Let's see if this rhombus is a square or not. We'll need to see if the adjacent sides are perpendicular. For that we'll need the slope.

Let's find the slope of AB.

$A = (x_1,y_1) = (3,5) \text{ and } B = (x_2,y_2) = (7,6)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{6 - 5}{7 - 3}\\\\m = \frac{1}{4}$

Segment AB has a slope of 1/4.

Do the same for BC

$B = (x_1,y_1) = (7,6) \text{ and } C = (x_2,y_2) = (6,2)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{2 - 6}{6 - 7}\\\\m = \frac{-4}{-1}\\\\m = 4$

Unfortunately the two slopes of 1/4 and 4 are not negative reciprocals of one another. One slope has to be negative while the other is positive, if we wanted perpendicular lines. Also recall that perpendicular slopes must multiply to -1.

We don't have perpendicular lines, so the interior angles are not 90 degrees each.

Therefore, this figure is not a rectangle and by extension it's not a square either.

The best description for this figure is a rhombus.

Answer 2

Answer:

  Rhombus

Step-by-step explanation:

The diagonals of a parallelogram bisect each other. That is, they have the same midpoint. If they are the same length, that parallelogram is a rectangle. If they cross at right angles, it is a rhombus. If it is a rectangle and rhombus, then it is a square.

Diagonal midpoints

The midpoints of AC and BD are ...

  (A+C)/2 and (B+D)/2

To determine if the midpoints are the same, we can skip the division by 2 and simply look at the sums:

  A +C = (3, 5) +(6, 2) = (9, 7)

  B +D = (7, 6) +(2, 1) = (9, 7)

The midpoints of the diagonals are the same, so the figure is at least a parallelogram.

Diagonal vectors

The diagonal vectors will be the same length if the figure is a rectangle. They will be perpendicular if the figure is a rhombus. The vectors are ...

  AC = C -A = (6, 2) -(3, 5) = (3, -3)

  BD = D -B = (2, 1) -(7, 6) = (-5, -5)

The length of each of these is the root of the sum of squares of its components. These are obviously different lengths (3√2 vs 5√2).

The dot-product of these will be zero if they are perpendicular:

  AC·BD = x1·x2 +y1·y2 = (3)(-5) +(-3)(-5) = -15 +15 = 0

Conclusion

The diagonals are different length and mutual perpendicular bisectors, so the figure is a rhombus.

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Additional comment

Looking at the dot-product is a simple way to check that the slopes are opposite reciprocals. The slope of a vector with components (x, y) is m = y/x.

The requirement that slopes be opposite reciprocals means ...

  y1/x1 = -1/(y2/x2) . . . . . . . . slope relationship

  (y1)(y2)/((x1)(x2)) = -1 . . . . . multiply by y2/x2

  y1·y2 = -x1·x2 . . . . . . . . . . multiply by (x1·x2)

  x1·x2 +y1·y2 = 0 . . . . . . . . add x1·x2

This shows the vector dot product being zero is equivalent to the slopes being opposite reciprocals. The vectors are perpendicular in this case.