Question

The coordinates of the vertices of quadrilateral ABCD are A(-5, 1), B(-2,5), C(5, 3),

Answer 1

Step-by-step explanation:

To calculate a slope, we need to apply: $m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Applying formula to each slope:

$m_{AB}=\frac{5-1}{-2-(-5)} = \frac{4}{3} \\m_{BC}=\frac{3-5}{5-(-2)} =\frac{-2}{7}\\m_{CD}=\frac{-1-3}{2-5}=\frac{-4}{-3} =\frac{4}{3}$

So, as you can see, there are equal pair of slopes, meaning that they are parallels, which demonstrate that it's actually a quadrilateral figure.

Answer 2

Answer: slope (AB)= 4/3 slope(BC)=-2/7 slope(CD) =4/3 slope(AD) = -2/7

Step-by-step explanation:

To find the slope, lets take a pair ones after the other

A(-5, 1) B(-2,5)

Given; x₁ =-5 y₁ = 1 x₂=-2 y₂=5

slope(AB) = y₂ - y₁ / x₂ - x₁

=5 - 1 / -2--5

=4/3

slope BC

B(-2, 5) c(5,3)

Given;

x₁=-2 y₁=5 x₂=5 y₂=3

slope(BC) = y₂ - y₁ / x₂ - x₁

= 3-5 / 5--2

=-2/7

slope CD

C(5, 3) D(2, -1)

Given;

x₁ = 5 y₁ = 3 x₂ = 2 y₂= -1

slope(CD)= y₂ - y₁ / x₂ - x₁

= -1-3 / 2-5

= -4/-3

= 4/3

slope AD

A(-5, 1) D(2, -1)

Given;

x₁ = -5 y₁ =1 x₂ =2 y₂ =-1

slope(AD) = y₂ - y₁ / x₂ - x₁

= -1-1 / 2--5

= -2/7