A quadrilateral with two pairs of parallel sides and with diagonals that are perpendicular is a rhombus

Question

A quadrilateral with two pairs of parallel sides and with diagonals that are perpendicular is a rhombus. Quadrilateral ABCD has vertices A(-2,2), B(1,6), C(6,6), D(3,2). Show that ABCD is a rhombus.

 

 

How do you show that. I've looked it up trying to find similar questions as to this one but nothing came up. 

Answer 1

ABCD is a rhombus

BD and AC are the diagonals that are perpendicular

find the slope of each diagonal

perpendicular lines have slopes that are negative reciprocals of each other

slope of AC is...

(6-2)/(6-[-2])=4/8=1/2

slope of BD is...

(6-2)/(1-3)=4/-2=-2

1/2 and -2 are negative reciprocals of each other

now show that AB and DC have the same slope and therefore are parallel

AB...

(6-2)/(1-[-2])=4/3

DC...

(6-2)/(6-3)=4/3

you can find the slopes of AD and BC and they are 0/5=0, so they are horizontal segments

now show that AB and DC are equal in length using the distance formula

AB...

√(6-2)²+(1-[-2])²=√(16+9)=5

DC...

√(6-2)²+(6-3)²=√(16+9)=5

this should suffice

if you want to find the lengths of AD and BC just use the distance formula again to get 5 and 5