Question

Determine whether quadrilateral ABCD with vertices A(–4, –5), B(–3, 0), C(0, 2), and D(5, 1) is a trapezoid.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

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:

$\dfrac{d-b}{c-a}$

So, the slope of AB is:

$\dfrac{0-(-5)}{-3-(-4)}\\\\=\dfrac{5}{-3+4}\\\\=\dfrac{5}{1}=5$

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:

$\dfrac{2-1}{0-5}\\\\=\dfrac{1}{-5}=-\dfrac{1}{5}$

Hence slope AB=-1/5.

Step 3:

slope of BC with vertices B(-3,0) and C(0,2) is:

$\dfrac{2-0}{0-(-3)}\\\\=\dfrac{2}{3}$

Step 4:

slope of AD with vertices A(-4,-5) and D(5,1) is:

$=\dfrac{1-(-5)}{5-(-4)}\\\\=\dfrac{1+5}{5+4}\\\\=\dfrac{6}{9}\\\\=\dfrac{2}{3}$

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)