Clayton needs to reflect the triangle below across the line y = x. On a coordinate plane, triangle A B C has points (5.5, 7), (6, 2), (4, 4).

Which statements about the reflection are true? Check all that apply.
Clayton could use the relationship (x, y) right-arrow (y, x) to find the points of the image.
Clayton could negate both the x and y values in the points to find the points of the image.
C’ will remain in the same location as C because it is on the line of reflection.
C’ will move because all points move in a reflection.
The image and the pre-image will be congruent triangles.
The image and pre-image will not have the same orientation because reflections flip figures.

could you reply to this with the question so i can ?

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.