Question

triangle abc was transformed using the rule (x, y) → (-y, x). the vertices of the triangles are shown a(-1, 1) a(-1, -1) b(1, 1) b(-1, 1) c(1, 4) c(-4, 1) which best describes the transformation? the transformation was a 90° rotation about the origin the transformation was a 180° rotation about the origin the transformation was a 270° rotation about the origin the transformation was a 360° rotation about the origin

Explanation:

Step1: Recall rotation rules

• \(90^{\circ}\) rotation about the origin: \((x,y)\to(-y,x)\)
• \(180^{\circ}\) rotation about the origin: \((x,y)\to(-x,-y)\)
• \(270^{\circ}\) rotation about the origin: \((x,y)\to(y, -x)\) (equivalent to \((x,y)\to(-(-y),x)\) when considering the general form of rotation matrices and composition of rotations)
• \(360^{\circ}\) rotation about the origin: \((x,y)\to(x,y)\)

Step2: Check the given rule \((x,y)\to(-y,x)\)

• For a \(270^{\circ}\) rotation about the origin, if we start with the standard rotation matrix. A \(90^{\circ}\) rotation is \(

$$\begin{pmatrix}\cos90^{\circ}&-\sin90^{\circ}\\\sin90^{\circ}&\cos90^{\circ}\end{pmatrix}$$

=

$$\begin{pmatrix}0& - 1\\1&0\end{pmatrix}$$

\), a \(270^{\circ}\) rotation is \(

$$\begin{pmatrix}\cos270^{\circ}&-\sin270^{\circ}\\\sin270^{\circ}&\cos270^{\circ}\end{pmatrix}$$

=

$$\begin{pmatrix}0&1\\ - 1&0\end{pmatrix}$$

\). If we have a point \((x,y)\), applying the transformation \((x,y)\to(y,-x)\) (which is the same as \((x,y)\to(-(-y),x)\)).

• Let's take a general point \((x,y)\). After a \(270^{\circ}\) rotation about the origin, using the rule \((x,y)\to(y, - x)\) (or rewritten as \((x,y)\to(-(-y),x)\) which matches the given rule \((x,y)\to(-y,x)\) when we consider the property of negative signs and rotation directions).

So the transformation was a \(270^{\circ}\) rotation about the origin.

Answer:

The transformation was a \(270^{\circ}\) rotation about the origin.