Quadrilateral ABCD is translated to get quadrilateral A'B'C'D'. vertex A is at (-5,2) , and vertex A' is ay (2,-2), quadrilateral ABCD is translated blank if vertex B is at (-6,5), then vertex B is at

(-1,3)

Step-by-step explanation:

Given that the quadrilateral ABCD is translated to get quadrilateral A'B'C'D'. The vertex A is at (-5,2) is translated to vertex A' is at (2,-2).

The translation is x-direction, a, is the difference of x-coordinates of both the points, a=2-(-5)=2+5=7.

The translation is y-direction, b, is the difference of y-coordinates of both the points, b = -2-2 = -4.

As all the points are translated by the same magnitude as well as in the same direction, so by adding a to x and b to y coordinate of any points, the translated point can be determined.

The vertex of point B is (-6,5), so, the vertex of the point B' would be (-6+a, 5+b)=(-6+7, 5+(-4))=(-1,3).

Hence, the vertex of point B' is (-1,3).

∠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.