Polygon ABCDE has the vertices A(2, 8), B(4, 12), C(10, 12), D(8, 8), and E(6, 6). Polygon MNOPQ has the vertices M(-2, 8), N(-4, 12), O(-10, 12), P(-8, 8), and Q(-6, 6).
A transformation or sequence of transformations that can be performed on polygon ABCDE to show that it is congruent to polygon MNOPQ is a .

If polygon MNOPQ is translated 3 units right and 5 units down, it will coincide with a congruent polygon, VWXYZ, with its vertices at .

We can see that the polygon MNOPQ is a reflection of the polygon ABCDE across the y-axis (only the abscissas change their signs from positive to negative) .

MrRoyal MrRoyal · Answer 2 · 2022-03-21T15:43:29+01:00

A transformation or sequence of transformations that can be performed on polygon ABCDE to show that it is congruent to polygon MNOPQ is a reflection across the y-axis

If polygon MNOPQ is translated 3 units right and 5 units down, it will coincide with a congruent polygon, VWXYZ, with its vertices at V(5, 3), B(7, 5), C(13, 7), D(11, 3), and E(9, 1).

The vertices of polygon ABCDE are given as:

A(2, 8), B(4, 12), C(10, 12), D(8, 8), and E(6, 6).

The vertices of polygon MNOPQ are given as:

M(-2, 8), N(-4, 12), O(-10, 12), P(-8, 8), and Q(-6, 6).

Notice that the x-coordinates of the vertices of polygon MNOPQ are opposites of x-coordinates of the vertices of polygon ABCDE, while the y-coordinates remain the same.

This means that the transformation from ABCDE to MNOPQ is a reflection across the y-axis

When the vertices of polygon ABCDE are shifted 3 units right and 5 units down, the follow the rule:

[tex](x,y) \to (x + 3,y-5)[/tex]

So, we have:

V(5, 3), B(7, 5), C(13, 7), D(11, 3), and E(9, 1).