How can you use distance formula and/or slope to classify a quadrilateral as a trapezoid?

a

step-by-step explanation:

given the 2 equations

- x + y = - 2 → (1)

2x - 3y = 4 → (2)

multiplying (1) by 2 and adding to (2) will eliminate the x- term

- 2x + 2y = - 4 → (3)

add (2) and (3) term by term

(2x - 2x) + (- 3y + 2y) = (4 - 4), simplifying gives

- y = 0 ⇒ y = 0

substitute y = 0 in either of the 2 equations for corresponding value of x

substituting in (2) gives

2x = 4 ( divide both sides by 2 )

x = 2

solution is (2, 0 ) → a

20/27

​​

step-by-step explanation:

1 simplify brackets

-\frac{8}{9}\times \frac{5}{6}−

​9

​

​8

​​ ×

​6

​

​5

​​

2 use this rule: \frac{a}{b}\times \frac{c}{d}=\frac{ac}{bd}

​b

​

​a

​​ ×

​d

​

​c

​​ =

​bd

​

​ac

​​

-\frac{8\times 5}{9\times 6}−

​9×6

​

​8×5

​​

3 simplify 8\times 58×5 to 4040

-\frac{40}{9\times 6}−

​9×6

​

​40

​​

4 simplify 9\times 69×6 to 5454

-\frac{40}{54}−

​54

​

​40

​​

5 simplify \frac{40}{54}

​54

​

​40

​​   to \frac{20}{27}

​27

​

​20

​​