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Quadrilateral ABCD have vertices at A (-4,4), B(1, 1,), C(4,6), and D (-1,9). Based on the properties of the diagonals, I’d quadrilateral ABCD a rectangle, rhombus, or square? Use the distance and slope formulas to prove your conclusion.

Quadrilateral ABCD have vertices at A 44 B1 1 C46 and D 19 Based on the properties of the diagonals Id quadrilateral ABCD a rectangle rhombus or square Use the class=

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Quadrilateral ABCD with vertices at A (-4,4), B(1, 1,), C(4,6), and D (-1,9). a rhombus.

Quadrilateral

A quadrilateral is a polygon with four sides and four angles. A rhombus is a quadrilateral in which opposite sides are equal and parallel. All the sides of a rhombus are congruent to each other, and opposite angles are congruent.

From the plot of Quadrilateral ABCD, we can see that all sides are equal. Quadrilateral ABCD with vertices at A (-4,4), B(1, 1,), C(4,6), and D (-1,9). a rhombus.

Find out more on Quadrilateral at: https://brainly.com/question/23935806

Answer:

Quadrilateral is a square.

Step-by-step explanation:

NOT 100% sure answer is correct.

Slope of AB             Slope of CD          Slope of BC            Slope of AD

m = [tex]\frac{y2 - y1}{x2 - x1}[/tex]                   m = [tex]\frac{y2 - y1}{x2 - x1}[/tex]             m = [tex]\frac{y2 - y1}{x2 - x1}[/tex]                  m = [tex]\frac{y2 - y1}{x2 - x1}[/tex]

ΔX = 1 - -4 = 5       ΔX = -1 – 4 = -5       ΔX = 1 – 4 = -3      ΔX = -1 - -4 = 3

ΔY = 1 – 4 = -3      ΔY = 9 – 6 = 3        ΔY = 1 – 6 = -5       ΔY = 9 – 4 = 5

m = -3/5 = -0.6       m = -3/5 = -0.6         m = 5/3 = 1.6           m = 5/3 = 1.6    

           

Distance AB                                           Distance CD

d = [tex]\sqrt{(x2-x1)^{2} + (y2-y1)^{2}[/tex]                 d = [tex]\sqrt{(x2-x1)^{2} + (y2-y1)^{2}[/tex]

(x2 – x1) = (1 - -4) = 5                            (x2 – x1) = (-1 – 4) = -5

(y2 – y1) = (1 – 4) = -3                          (y2 – y1) = (9 - 6) = 3

Square results & sum them up:          Square results & sum them up:

[tex](5)^{2}[/tex] + [tex](-3)^{2}[/tex] = 25 + 9 = 34                    [tex](-5)^{2}[/tex] + [tex](3)^{2}[/tex] = 25 + 9 = 34

Find square root: √34 or 5.8   Find square root: √34 or 5.8

Distance BC                                            Distance AD

d = [tex]\sqrt{(x2-x1)^{2} + (y2-y1)^{2}[/tex]                  d = [tex]\sqrt{(x2-x1)^{2} + (y2-y1)^{2}[/tex]

(x2 – x1) = (1 – 4) = -3                             (x2 – x1) = (-1 - -4) = 3

(y2 – y1) = (1 – 6) = -5                              (y2 – y1) = (9 – 4) = 5

Square results & sum them up:          Square results & sum them up:

[tex](-3)^{2}[/tex] + [tex](-5)^{2}[/tex] = 9 + 25 = 34                   [tex](3)^{2}[/tex] + [tex](5)^{2}[/tex] = 9 + 25 = 34

Find square root: √34 or 5.8          Find square root: √34 or 5.8

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