Question 704144: If someone could help me with this question, I would really appreciate it. Points E (0,5), F (4,2), G (0, -1) and H (-4,2) are the vertices of a quadrilateral.
a)Verify that quadrilateral EFGH is a rhombus.
I know that I have to show all four sides are congruent...but would this just be using the distance formula? Also, I'm pretty sure there is more to a rhombus than just the sides being congruent.
b)Verify that the diagonals of quadrilateral EFGH are perpendicular bisectors.
I have no idea for this one!
Thank you so much in advance!
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! Opposite sides are parallel, and
the diagonals are perpendicular bisectors of each other.
You can prove it is a rhombus if you prove that
the diagonals are perpendicular bisectors of each other.
It is easy to see that EG is vertical, part of the line  ,
and that FH is horizontal, part of the line  ,
and that the diagonals intersect at point (0,2).
What remains to prove is that (0,2) is the midpoint of EG, and the midpoint of FH.
You do remember that the coordinates of the midpoint of a segment are the averages of the coordinates of the two end points, right?
|
|
|
