Which rule describes the composition of transformations that maps rectangle PQRS to P''Q''R''S''? R0,270° ∘ T0,2(x, y) R0,180° ∘ T2,0 (x, y) T0,2 ∘ R0,270°(x, y) R0,2 ∘ T0,180°(x, y)

Answer:  The correct option is (C) T0,2(x, y)∘ R0, 270°.Step-by-step explanation:  We are given to select the rule that describes the composition of transformations that maps rectangle PQRS to P''Q''R''S''.From the graph, we note thatthe co-ordinates of the vertices of rectangle PQRS are P(-3, -5), Q(-2, -5), R(-2, -1) and S(-3, -1).and the co-ordinates of the vertices of rectangle P''Q''R''S'' are P''(-5, 5), Q''(-5, 4), R''(-1, 4) and S(-1, 5).Since the rectangle PQRS lies in the Quadrant III and P''Q''R''S'' lies in Quadrant II, so the rotation will be either 90° clockwise or 270° counterclockwise about the origin.In the given options, we do not have rotation of 90°, so will consider the rotation through 270° counterclockwise about the origin.Now, after this rotation, the vertices will transform according to the rule  (x, y)   ⇒   (y, -x).Therefore, the vertices of the image rectangle P'Q'R'S', after rotation through  270° counterclockwise about the origin, becomesP(-3, -5)    ⇒  P'(-5, 3), Q(-2, -5)   ⇒  Q'(-5, 2),R(-2, -1)     ⇒  R'(-1, 2),S(-3, -1)     ⇒ S'(-1, 3).Now, to make the vertices of P'Q'R'S' coincide with the vertices of P''Q''R"S", we need to add 2 units to the y co-ordinate of each vertex , so thatP'(-5, 3)    ⇒   P''(-5, 3+2) = P''(-5, 5), Q'(-5, 2)   ⇒   Q"(-5, 2+2) = Q''(-5, 4),R'(-1, 2)     ⇒   R''(-1, 2+2) = R"(-1, 4),S'(-1, 3)     ⇒   S''(-1, 3+2) = S''(-1, 5).Thus, the required transformation rule isrotation through 270° counterclockwise about the origin and a translation of (x, y)   ⇒  (x, y+2).Since the rigid transformations are written from right to left, so option (C) is correct.