Question

Triangle MNO and triangle PQR are shown on the coordinate plane. A triangle on the coordinate plane with vertices M at 1 comma 1, N at 3 comma 1, and O at 3 comma 4. A second triangle with vertices P at negative 2 comma negative 2, Q at negative 6 comma negative 2, and R at negative 6 comma negative 8. What sequence of transformations proves ΔMNO ~ ΔPQR? ΔMNO was rotated 180° clockwise about the origin, then dilated by a factor of 2 about the origin to form ΔPQR. ΔMNO was translated up 6 units, then dilated by a factor of 2 about the origin to form ΔPQR. ΔMNO was reflected across the x-axis, then dilated by a factor of 3 about the origin to form ΔPQR. ΔMNO was reflected across the y-axis, then dilated by a factor of 3 about the origin to form ΔPQR.

Answer 1

The correct sequence of transformations is the first one: ΔMNO was rotated 180° clockwise about the origin, then dilated by a factor of 2 about the origin to form ΔPQR.

Triangle MNO is initially at M(1, 1), N(3, 1), and O(3, 4).

Triangle PQR is initially at P(-2, -2), Q(-6, -2), and R(-6, -8).

Now, let's go through each sequence of transformations:

Rotation 180° clockwise about the origin, then dilated by a factor of 2 about the origin: Rotate ΔMNO 180° clockwise about the origin: The new coordinates will be M'(1, 1) → (-1, -1), N'(3, 1) → (-3, -1), O'(3, 4) → (-3, -4).

Dilate the rotated triangle by a factor of 2 about the origin: The new coordinates will be M''(-1, -1) → (-2, -2), N''(-3, -1) → (-6, -2), O''(-3, -4) → (-6, -8).

This sequence of transformations matches the coordinates of ΔPQR, so ΔMNO ~ ΔPQR.