Answer:
We are given a quadrilateral ABCD with vertices A(–4, –5), B(–3, 0), C(0, 2), and D(5, 1).
<u>Step 1:</u>
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:
![\dfrac{d-b}{c-a}](https://tex.z-dn.net/?f=%5Cdfrac%7Bd-b%7D%7Bc-a%7D)
So, the slope of AB is:
![\dfrac{0-(-5)}{-3-(-4)}\\\\=\dfrac{5}{-3+4}\\\\=\dfrac{5}{1}=5](https://tex.z-dn.net/?f=%5Cdfrac%7B0-%28-5%29%7D%7B-3-%28-4%29%7D%5C%5C%5C%5C%3D%5Cdfrac%7B5%7D%7B-3%2B4%7D%5C%5C%5C%5C%3D%5Cdfrac%7B5%7D%7B1%7D%3D5)
Hence slope AB=5.
<u>Step 2:</u>
Now we have to find the slope of DC.
with vertices D(5,1) and C(0,2)
So, the slope of DC is:
![\dfrac{2-1}{0-5}\\\\=\dfrac{1}{-5}=-\dfrac{1}{5}](https://tex.z-dn.net/?f=%5Cdfrac%7B2-1%7D%7B0-5%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B-5%7D%3D-%5Cdfrac%7B1%7D%7B5%7D)
Hence slope AB=-1/5.
<u>Step 3:</u>
slope of BC with vertices B(-3,0) and C(0,2) is:
![\dfrac{2-0}{0-(-3)}\\\\=\dfrac{2}{3}](https://tex.z-dn.net/?f=%5Cdfrac%7B2-0%7D%7B0-%28-3%29%7D%5C%5C%5C%5C%3D%5Cdfrac%7B2%7D%7B3%7D)
<u>Step 4:</u>
slope of AD with vertices A(-4,-5) and D(5,1) is:
![=\dfrac{1-(-5)}{5-(-4)}\\\\=\dfrac{1+5}{5+4}\\\\=\dfrac{6}{9}\\\\=\dfrac{2}{3}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B1-%28-5%29%7D%7B5-%28-4%29%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%2B5%7D%7B5%2B4%7D%5C%5C%5C%5C%3D%5Cdfrac%7B6%7D%7B9%7D%5C%5C%5C%5C%3D%5Cdfrac%7B2%7D%7B3%7D)
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)