Question

A quadrilateral has vertices at A (-5, 5), B (1, 8), C (4, 2), and D (-2, -2). Use slope to determine if the quadrilateral is a rectangle. Show your work. (Try to use point slope form)

Answer 1

Answer:

not a rectangle

Step-by-step explanation:

There are several ways to determine whether the quadrilateral is a rectangle. Computing slope is one of the more time-consuming. We can already learn that the figure is not a rectangle by seeing if the midpoint of AC is the same as that of BD. (It is not.) A+C = (-5+4, 5+2) = (-1, 7). B+D = (1-2, 8-2) = (-1, 6). (A+C)/2 ≠ (B+D)/2, so the midpoints of the diagonals are different points.

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The slope of AB is ∆y/∆x, where the ∆y is the change in y-coordinates, and ∆x is the change in x-coordinates.

... AB slope = (8-5)/(1-(-5)) = 3/6 = 1/2

The slope of AD is computed in similar fashion.

... AD slope = (-2-5)/(-2-(-5)) = -7/3

The product of these slopes is (1/2)(-7/3) = -7/6 ≠ -1. Since the product is not -1, the segments AB and AD are not perpendicular to each other. Adjacent sides of a rectangle are perpendicular, so this figure is not a rectangle.

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Our preliminary work with the diagonals showed us the figure was not a parallelogram (hence not a rectangle). For our slope calculation, we "magically" chose two sides that were not perpendicular. In fact, this choice was by "trial and error". Side BC is perpendicular to AB, so we needed to choose a different side to find one that wasn't. A graph of the points is informative, but we didn't start with that.