Question

which transformation carries point h(1, 4) to point k(-4, -1)?
a counterclockwise rotation of 90° about the origin
a clockwise rotation of 180° about the origin
a counterclockwise rotation of 270° about the origin
a clockwise rotation of 360° about the origin

Explanation:

Step1: Recall rotation rules

The rotation rules about the origin are: For a point $(x,y)$ rotated counter - clockwise by $90^{\circ}$, the new point is $(-y,x)$; rotated clockwise by $180^{\circ}$, the new point is $(-x,-y)$; rotated counter - clockwise by $270^{\circ}$ (same as clockwise by $90^{\circ}$), the new point is $(y, - x)$; rotated clockwise by $360^{\circ}$, the point remains the same $(x,y)$.

Step2: Apply rules to point $H(1,4)$

For a counter - clockwise rotation of $90^{\circ}$ about the origin, if $(x = 1,y = 4)$, the new point is $(-4,1)$.
For a clockwise rotation of $180^{\circ}$ about the origin, if $(x = 1,y = 4)$, the new point is $(-1,-4)$.
For a counter - clockwise rotation of $270^{\circ}$ about the origin (equivalent to a clockwise rotation of $90^{\circ}$), if $(x = 1,y = 4)$, the new point is $(4,-1)$.
For a clockwise rotation of $360^{\circ}$ about the origin, if $(x = 1,y = 4)$, the new point is $(1,4)$.
For a counter - clockwise rotation of $270^{\circ}$ about the origin, for point $H(1,4)$, we swap the $x$ and $y$ coordinates and change the sign of the new $x$ - coordinate. So $(1,4)$ becomes $(4,-1)$. If we consider the general rotation matrix for a counter - clockwise rotation of an angle $\theta$ about the origin

$$\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$

, for $\theta = 270^{\circ}$, $\cos(270^{\circ})=0,\sin(270^{\circ})=- 1$. Applying the matrix

$$\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}1\\4\end{pmatrix}=\begin{pmatrix}4\\-1\end{pmatrix}$$

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Answer:

a counterclockwise rotation of 270° about the origin