Rhombus PQRS with vertices P(-8, 6), Q(-4, 8), R(0, 6), and S(-4, 4) REFLECTED over the line y = 2.

P'(-8, -2), Q'(-4, -4), R'(0, -2) and  S'(-4, 0)

Step-by-step explanation:

Transformation is the movement of a point from its initial location to a new location. Types of transformation are reflection, rotation, translation and dilation.

Reflection is the flipping of a point about an axis.

Given the Rhombus PQRS with vertices P(-8, 6), Q(-4, 8), R(0, 6), and S(-4, 4), if the rhombus is reflected over the y = 2 line the new point is:

For P(-8, 6): The point P has a y coordinate 4 units above the y = 2 line. Hence if it is reflected over the line y = 2, the new point would have a y coordinate 4 units below the y = 2 line which is P'(-8, -2)

For Q(-4, 8): The point Q has a y coordinate 6 units above the y = 2 line. Hence if it is reflected over the line y = 2, the new point would have a y coordinate 6 units below the y = 2 line which is Q'(-4, -4)

For R(0, 6): The point R has a y coordinate 4 units above the y = 2 line. Hence if it is reflected over the line y = 2, the new point would have a y coordinate 4 units below the y = 2 line which is R'(0, -2)

For S(-4, 4): The point S has a y coordinate 2 units above the y = 2 line. Hence if it is reflected over the line y = 2, the new point would have a y coordinate 2 units below the y = 2 line which is S'(-4, 0)

Therefore the reflected rhombus is at P'(-8, -2), Q'(-4, -4), R'(0, -2) and  S'(-4, 0)

something over 10 so ? /10

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)