Respuesta :
Answer:
[tex]|F'H'| = 2 * |FH|[/tex]
Step-by-step explanation:
Given
[tex]E = (0,1)[/tex] [tex]E' = (-1,2)[/tex]
[tex]F = (1,1)[/tex] [tex]F' = (1,2)[/tex]
[tex]G = (2,0)[/tex] [tex]G' =(3,0)[/tex]
[tex]H = (0,0)[/tex] [tex]H' = (-1,0)[/tex]
[tex](x,y) = (1,0)[/tex] -- center
[tex]k = 2[/tex] --- scale factor
See comment for proper format of question
Required
Compare FH to F'H'
From the question, we understand that the scale of dilation from EFGH to E'F'G'H is 2;
Irrespective of the center of dilation, the distance between corresponding segment will maintain the scale of dilation.
i.e.
[tex]|F'H'| = k * |FH|[/tex]
[tex]|F'H'| = 2 * |FH|[/tex]
To prove this;
Calculate distance of segments FH and F'H' using:
[tex]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]
Given that:
[tex]F = (1,1)[/tex] [tex]F' = (1,2)[/tex]
[tex]H = (0,0)[/tex] [tex]H' = (-1,0)[/tex]
We have:
[tex]FH = \sqrt{(1- 0)^2 + (1- 0)^2}[/tex]
[tex]FH = \sqrt{(1)^2 + (1)^2}[/tex]
[tex]FH = \sqrt{1 + 1}[/tex]
[tex]FH = \sqrt{2}[/tex]
Similarly;
[tex]F'H' = \sqrt{(1 --1)^2 + (2 -0)^2}[/tex]
[tex]F'H' = \sqrt{(2)^2 + (2)^2}[/tex]
Distribute
[tex]F'H' = \sqrt{(2)^2(1 +1)}[/tex]
[tex]F'H' = \sqrt{(2)^2*2}[/tex]
Split
[tex]F'H' = \sqrt{(2)^2} *\sqrt{2}[/tex]
[tex]F'H' = 2 *\sqrt{2}[/tex]
[tex]F'H' = 2\sqrt{2}[/tex]
Recall that:
[tex]|F'H'| = 2 * |FH|[/tex]
So, we have:
[tex]2\sqrt 2 = 2 * \sqrt 2[/tex]
[tex]2\sqrt 2 = 2\sqrt 2[/tex] --- true
Hence, the dilation relationship between FH and F'H' is::
[tex]|F'H'| = 2 * |FH|[/tex]
Answer:NOTT !! A segment in the image has the same length as its corresponding segment in the pre-image.
Step-by-step explanation:
