Question

A quadrilateral ABCD that has vertices A(2, 3), B(9, 5), C(4, 0) and D(-3, -2). What
kind of quadrilateral is ABCD?

Answer 1

Step-by-step explanation:

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Answer 2

ABCD is a parallelogram.

Step-by-step explanation:

A=(2,3)

B=(9,5)

C=(4,0)

D=(-3,-2)

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

Using the formula

$AB=\sqrt{(9-2)^2+(5-3)^2}=\sqrt{7^2+4}=\sqrt{49+4}=\sqrt{53}$units

$BC=\sqrt{(4-9)^2+(0-5)^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt 2$units

$CD=\sqrt{(-3-4)^2+(-2-0)^2}=\sqrt{(-7)^2+(-2)^2}=\sqrt{49+4}=\sqrt{53}$units

$AD=\sqrt{(-3-2)^2+(-2-3)^2}=\sqrt{(-5)^2+(-5)^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt 2$units

AB=CD,AD=BC

When opposite sides of quadrilateral are equal then, the quadrilateral is parallelogram or rectangle.

But in rectangle each angle is equal to 90 degrees.

In attached figure, each angle is not equal to 90 degrees.

Hence, ABCD is a parallelogram.

#Learns more:

https://brainly.in/question/728617:answered by brainly user.