Respuesta :
Answer:
We are given a quadrilateral ABCD with vertices A(–4, –5), B(–3, 0), C(0, 2), and D(5, 1).
Step 1:
We find the slope of side AB.
with vertices A(-4,-5) and B(-3,0)
Hence, the slope of two points (a,b) and (c,d) is calculates as:
[tex]\dfrac{d-b}{c-a}[/tex]
So, the slope of AB is:
[tex]\dfrac{0-(-5)}{-3-(-4)}\\\\=\dfrac{5}{-3+4}\\\\=\dfrac{5}{1}=5[/tex]
Hence slope AB=5.
Step 2:
Now we have to find the slope of DC.
with vertices D(5,1) and C(0,2)
So, the slope of DC is:
[tex]\dfrac{2-1}{0-5}\\\\=\dfrac{1}{-5}=-\dfrac{1}{5}[/tex]
Hence slope AB=-1/5.
Step 3:
slope of BC with vertices B(-3,0) and C(0,2) is:
[tex]\dfrac{2-0}{0-(-3)}\\\\=\dfrac{2}{3}[/tex]
Step 4:
slope of AD with vertices A(-4,-5) and D(5,1) is:
[tex]=\dfrac{1-(-5)}{5-(-4)}\\\\=\dfrac{1+5}{5+4}\\\\=\dfrac{6}{9}\\\\=\dfrac{2}{3}[/tex]
As we know that parallel sides have same slope.
As slope of BC=AD.
Hence AD is parallel to BC.
and the slope of two opposite sides are not equal hence the two sides are not parallel.
Hence, the given quadrilateral ABCD is a trapezoid. ( since it has a pair of opposite sides which are parallel and a pair of non-parallel opposite sides)
