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Quadrilateral JKLM has vertices J(−2,4), K(4,1), L(2,−3), and M(−4,0). What statements describe JKLM? Select all that apply.

Multiple select question.

A)
The diagonals of quadrilateral JKLM are perpendicular.


B)
Quadrilateral JKLM is a trapezoid.


C)
Quadrilateral JKLM is a square.


D)
Quadrilateral JKLM is a parallelogram.


E)
The diagonals of quadrilateral JKLM bisect each other.


F)
The diagonals of quadrilateral JKLM are congruent.


G)
Quadrilateral JKLM is a rectangle.

1 Answer

2 votes

Answer:

D, E, F, G

Explanation:

You want to know which statements can describe quadrilateral JKLM, which has vertices J(−2, 4), K(4, 1), L(2, −3), and M(−4, 0).

Diagonals

We can check to see if the diagonals have the same midpoint by adding the coordinates of their ends:

J+L = (-2, 4) +(2, -3) = (-2+2, 4-3) = (0, 1)

K+M = (4, 1) +(-4, 0) = (4-4, 1+0) = (0, 1)

The diagonals have the same midpoint, so the diagonals bisect each other, and the figure is a parallelogram. .

Rectangle

We can check to see if the diagonals are the same length by looking at the differences of their end points:

J -L = (-2, 4) -(2, -3) = (-2-2, 4+3) = (-4, 7)

length of JL is √((-4)² +7²) = √65

K -M = (4, 1) -(-4, 0) = (4+4, 1-0) = (8, 1)

length of KM is √(8² +1²) = √65

The diagonals are congruent, so the figure is a rectangle.

The endpoint differences are not opposite reciprocals of each other, so the diagonals are not perpendicular. This means the rectangle is not a square.

Summary

The applicable statements are ...

  • D — JKLM is a parallelogram
  • E — the diagonals bisect each other
  • F — the diagonals are congruent
  • G — JKLM is a rectangle

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Additional comment

The diagonals will be perpendicular if the differences of their end points indicate opposite reciprocal slopes. That is, if one difference is (a, b) and the other difference is (c, d), the diagonals are perpendicular if ac+bd=0. Here, that sum is -4·8 +7·1 = -25, not 0.

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Quadrilateral JKLM has vertices J(−2,4), K(4,1), L(2,−3), and M(−4,0). What statements-example-1
User Yaniv Inbar
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