The rule RO, 90° • T-1, 1(x, y) is applied to ΔBCD to produce ΔB"C"D". Point B" of the final image is at (–4, 1).

(–2, –5)

(0, –5)

(2, 3)

(5, 0)

Can anyone help? I think it is either (0,-5) or (2,3) but I am not sure whether to rotate clockwise, or counter-clockwise?

Verified Answer :

The transformation starts with T-1, 1(x, y) and ends with RO, 90° (counter-clockwise).

To find the original point, we work backwards.

We start with RO, -90° (clockwise):
\((x, y)\rightarrow(y, -x)\)
which gives
\((-4, 1)\rightarrow(1, 4)\)

We then do the reverse of T-1, 1(x, y), which is T1, -1(x, y):
\((x, y)\rightarrow(x+1, y-1)\)
which gives
\((1, 4)\rightarrow(2, 3)\)

so the answer is \((2, 3)\).

(2,3) is the answer hope that helps