Respuesta :
F(3b,2b) G(b,0)
midpoint of GF would be [(x1+x2)/2 , (y1+y2)/2]
= [4b/2 , 2b/2]
= [2b, b]
Therefore C is the answer.
midpoint of GF would be [(x1+x2)/2 , (y1+y2)/2]
= [4b/2 , 2b/2]
= [2b, b]
Therefore C is the answer.
Answer:
C)(2b,b)
Step-by-step explanation:
We are given that a quadrilateral EFGH whose coordinates are E(2b,b),F(3b,2b),G(b,0) and H(0,0).
We have to find the midpoint of GF.
Midpoint formula :
[tex]x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}[/tex]
We have [tex]x_2=3b,x_1=b,y_2=2b,y_1=0[/tex]
Using the formula
Then, we get
Midpoint of GF,[tex]x=\frac{3b+b}{2}=2b, y=\frac{2b+0}{2}=b[/tex]
Hence, the midpoint of GF=(2b,b)
Answer:C)(2b,b)