Question

What is the most precise name for quadrilateral ABCD with vertices A(-3,1) , B(-1,4) , C(5,4) , and D(3,1)?

Answer 1

The best and most correct answer among the choices provided by your question is the fourth choice or letter D.

We can imply from the given points that the figure is four-side or a quadrilateral.

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Answer 2

Answer:

ABCD is a parallelogram.

Step-by-step explanation:

Given quadrilateral ABCD with vertices A(-3,1), B(-1,4), C(5,4) and D(3,1)

we have to give the most precise name for quadrilateral ABCD.

Using distance formula

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AB=\sqrt{(-1+3)^2+(4-1)^2}=\sqrt{4+9}=\sqrt{13}units$

$BC=\sqrt{(5+1)^2+(4-4)^2}=\sqrt{36}=\sqrt{6}units$

$CD=\sqrt{(3-5)^2+(1-4)^2}=\sqrt{4+9}=\sqrt{13}units$

$DA=\sqrt{(-3-3)^2+(1-1)^2}=\sqrt{36}=\sqrt{6}units$

Opposite sides of quadrilateral ABCD are equal.

Now, we find the slope of the sides of quadrilateral

$\text{Slope of AB=}\frac{y_2-y_1}{x_2-x_1}=\frac{4-1}{-1+3}=\frac{3}{2}$

$\text{Slope of BC=}\frac{y_2-y_1}{x_2-x_1}=\frac{4-4}{5+1}=0$

$\text{Slope of CD=}\frac{y_2-y_1}{x_2-x_1}=\frac{1-4}{3-5}=\frac{3}{2}$

$\text{Slope of DA=}\frac{y_2-y_1}{x_2-x_1}=\frac{1-1}{-3-3}=0$

Slopes of opposite sides are equal.

⇒ The sides of opposite sides of quadrilateral are parallel and equal.

⇒ ABCD is a parallelogram.

Option b) is correct.