Question

Coordinate plane with triangles EFG and GHI with E at negative 6 comma 2, F at negative 2 comma 6, G at negative 2 comma 2, H at negative 2 comma 4, and I at 0 comma 2
Which set of transformations would prove ΔEFG ~ ΔGHI?

Translate ΔGHI by the rule (x − 2, y + 0), and reflect ΔG′H′I′ over x = −4.
Reflect ΔGHI over x = −2, and translate ΔG′H′I′ by the rule (x − 2, y + 0).
Translate ΔGHI by the rule (x + 0, y + 2), and dilate ΔG′H′I′ by a scale factor of 2 from point H.
Reflect ΔGHI over x = −2, and dilate ΔG′H′I′ by a scale factor of 2 from point G.

Answer 1

Answer:

The set of transformation would prove ΔEFG ~ ΔGHI is "Reflect ΔGHI

over x = −2, and dilate ΔG′H′I′ by a scale factor of 2 from point G" ⇒

last answer

Step-by-step explanation:

* Lets revise the reflection and dilation

- A reflection is a transformation where each point in a shape appears

 at an equal distance on the opposite side of the line of reflection

- Reflection doesn't change the size of the original figure

- A dilation is a transformation that produces an image that is the same

 shape as the original, but is a different size

- The dilation has a scale factor and center of dilation

- If the scale factor greater than 1, then the image will be larger

- If the scale factor between 0 and 1, then the image will be smaller

* Lets solve the problem

- The vertices of Δ EFG are:

# E = (-6 , 2)

# F = (-2 , 6)

# G = (-2 , 2)

- The vertices of Δ GHI are:

# I = (0 , 2)

# H = (-2 , 4)

# G = (-2 , 2)

- From the figure

∵ Point I is in the opposite side of the vertical line x = -2

∵ Point I in ΔGHI is corresponding to point E in Δ GFE

∴ G'H'I' is the image of Δ GHI by reflection across the line x = -2

∴ Its vertices must be

# I = (-4 , 2)

# H = (-2 , 4)

# G = (-2 , 2)

- Find the length of the side H'G' by subtracting the y-coordinates

 of points H' and G' and find the length of side FG by subtracting

 the y-coordinates of points F and G

∵ The length of the side H'G' is 2 units ⇒ (4 - 2 = 2)

∵ The length of the side FG is 4 units ⇒ (6 - 2 = 4)

∵ FG/H'G' = 4/2 = 2

∴ Δ G'H'I' dilated by scale factor 2 and center G to get Δ GFE

∴ There is a constant ratio between the sides of Δ G'H'I' and Δ DFE

- Triangles are similar if their corresponding sides are proportion

∴ Δ G'H'I' similar to Δ DFE

- Reflection doesn't change the size of the figure

∵ Δ G'H'I' is congruent to Δ GHI

∴ Δ GHI  similar to Δ DFE

* The set of transformation would prove ΔEFG ~ ΔGHI is "Reflect

 ΔGHI over x = −2, and dilate ΔG′H′I′ by a scale factor of 2 from

 point G"

Answer 2

Answer: D) Reflect ΔGHI over x = −2, and dilate ΔG′H′I′ by a scale factor of 2 from point G.