Question

90° Clockwise and Counterclockwise Rotations Tutorial
Creator of Tutorial: e.mccormick

Answer 1

Many translations can be shown by some form of (x,y) to a new (x,y) notation. This will go over 90 degree rotations and some explanations of the notation.

Answer 2

For ordered pair Cartesian coordinates, you \(\textbf{always}\) use \((x,y)\) as the order. This means that the point \((1,4)\) always has the x value of 1 and y value of 4.

However, you may see some odd uses of this notation when you do rotations, like this:
\((x,y)\rightarrow (y,-x)\)

Now, this does not mean that the entire concept of ordered pairs is changing! In fact, they want you to keep the idea of ordered pairs intact.

This notation actually means this: Take the point \(P_1\) \((x,y)\) and find the x and y values. Let us call these \(x_1\) and \(y_1\). Now make a new point \(P_2\) that is going to have its own x and y, say \(x_2\) and \(y_2\). With the translation (\(\rightarrow\)), \(x_2=y_1\) and \(y_2=-x_1\).

So if I had written all that out in the notation it would have been:
\((x_1=x,y_1=y)\rightarrow (x_2=y,y_2=-x)\)
But that is still long for a mathematician, so they use the shorter form.

Another name for \((x,y)\rightarrow (y,-x)\) is the 90 degree clockwise rotation. The arrow, in this case, means translation, but this particular translation is a rotation. (If it was a double arrow, \(\Rightarrow\), it would mean implies! Do not get your arrows confused.)

Here is a more graphical version of this:
https://www.desmos.com/calculator/ureqospxjh

On that page, I show how the purple box is rotated 90 degrees clockwise to be the red box or 90 degrees left to become the green box. I have matched up the order of the points in the tables to help show how they are translated.

The calculations like \(1\le y \le 4 \{1< x < 3\}\) are only there for shading the box. They have no bearing on the placement of the points.

In the tables it shows:

\(\quad\begin{array}{|c|c|c|}\hline \\
\color{purple}{\text{First/Purple}}
&
\color{red}{\text{Second/Red}}
&
\color{green}{\text{Third/Green}}
\\[.25em]
\quad\begin{array}{c|c}
x&\huge\color{purple}{\bullet }\normalsize y\\ \hline
1&1\\ \hline
1&4\\ \hline
3&4\\ \hline
3&1
\end{array}\quad
&
\quad
\begin{array}{c|c}
x&\huge\color{red}{\bullet }\normalsize y\\ \hline
1&-1\\ \hline
4&-1\\ \hline
4&-3\\ \hline
1&-3
\end{array}\quad
&
\quad\begin{array}{c|c}
x&\huge\color{green}{\bullet }\normalsize y\\ \hline
-1&1\\ \hline
-4&1\\ \hline
-4&3\\ \hline
-1&3
\end{array}\quad\\ \\ \hline
\end{array}\)

The first table to the second is the clockwise \(90^\circ\) rotation:
\((x,y)\rightarrow (y,-x)\)

The first table to the third is the counterclockwise \(90^\circ\) rotation:
\((x,y)\rightarrow (-y,x)\)

And, you could use these tables to find the \(180^\circ\) rotation! Do a little thinking on this topic and you should see that solution.

Answer 3

thanks