Answer:
The answer is:
[tex]Q' = (-3, 3)\\\\R' = (-5, -3)\\\\S' = (-2, -2)\\\\T' = (0, 1)\\\\[/tex]
Step-by-step explanation:
All vertices of its quadrilateral were co-ordinated;
[tex]T(-2, 1), \ Q(1, 3), \ S(0, -2), \ R(3, -3)[/tex] Therefore, we first undertake the operation correct: [tex]T(2, 0) \ QRST[/tex], there;
[tex]T(2, 0) \ QRST[/tex] [tex]\to T(-2 + 2, 1), \ Q(1 + 2, 3), \ S(0 + 2, -2), \ R(3 + 2, -3)[/tex]
[tex]T(2, 0) \ QRST[/tex][tex]\to T(0, 1), \ Q(3, 3), \ S(2, -2), \ R(5, -3)[/tex]
A preimage (x, y) will be the image in the next phase R-y-axis reflecting mostly on the y-axis (-x, y) and we've got it;
R y-axis [tex](T(0, 1), \ Q(3, 3), \ S(2, -2), \ R(5, -3)) = \ T'(0, 1), \ Q'(-3, 3), \ S'(-2, -2), \ R'(-5, -3)[/tex]From which we get [tex](Ry-axis \ T(2, 0)) \ (QRST = T'(0, 1), \ Q'(-3, 3), \ S'(-2, -2),\ R'(-5,-3).[/tex]