Question

A triangle has vertices at R(1, 1), S(–2, –4), and T(–3, –3). The triangle is transformed according to the rule R0, 270°. What are the coordinates of S'?

(–4, 2)
(–2, 4)
(2, 4)
(4, –2)

Answer 1

To apply the transformation \(R0, 270^\circ\) to point S(–2, –4), we need to follow these steps:

1. **Translate the point to the origin**: Since the transformation is \(R0\), which likely indicates rotation around the origin, we translate S to the origin by keeping the coordinates as they are since the origin is (0, 0).

2. **Rotate the point by \(270^\circ\)**: The rotation matrix for a counterclockwise rotation by \(270^\circ\) (which is equivalent to a clockwise rotation by \(90^\circ\)) is given by:
\[
R = \begin{pmatrix}
\cos 270^\circ & -\sin 270^\circ \\
\sin 270^\circ & \cos 270^\circ
\end{pmatrix}
=
\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}
\]

Applying this rotation matrix to point S(–2, –4):
\[
\begin{pmatrix}
x' \\
y'
\end{pmatrix}
=
\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}
\begin{pmatrix}
-2 \\
-4
\end{pmatrix}
=
\begin{pmatrix}
0 \cdot (-2) + 1 \cdot (-4) \\
-1 \cdot (-2) + 0 \cdot (-4)
\end{pmatrix}
=
\begin{pmatrix}
-4 \\
2
\end{pmatrix}
\]

Thus, the new coordinates of the transformed point S' are \((-4, 2)\).

Comparing with the provided options:
- (–4, 2) ✅
- (–2, 4)
- (2, 4)
- (4, –2)

Thus, the correct coordinates of \(S'\) after the transformation are \((-4, 2)\).