Question

rotate figure qrst 90° clockwise around the origin.

Explanation:

Step1: Identify original coordinates

First, find the coordinates of each vertex of figure \( QRST \). From the graph:
- \( Q(-4, -2) \)
- \( R(-1, -2) \) (Wait, correction: Looking at the grid, \( R \) is at \( (-1, -2) \)? Wait no, the grid lines: Let's re - check. The x - axis: from - 4, - 3, - 2, - 1, 0... Wait, the original figure: \( Q \) is at \( (-4, -2) \)? Wait no, the blue dots: Let's see, the horizontal line (x - axis) and vertical (y - axis). Let's list the coordinates properly:
Looking at the graph, the coordinates are:
- \( Q(-4, -2) \)? Wait no, the x - coordinate for \( Q \): the first blue dot on the left, x=-4? Wait, no, the grid has x from - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4. The y - axis: from - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4.
Wait, \( Q \): x=-4, y=-2? \( R \): x=-1, y=-2? \( S \): x=-1, y=-4? \( T \): x=-4, y=-4? Wait, no, let's check the positions:
\( Q \) is at \( (-4, -2) \), \( R \) is at \( (-1, -2) \), \( S \) is at \( (-1, -4) \), \( T \) is at \( (-4, -4) \).

Step2: Apply 90° clockwise rotation rule

The rule for rotating a point \((x,y)\) 90° clockwise about the origin is \((x,y)\to(y, - x)\).

- For point \( Q(-4, -2) \):
Using the rule \((x,y)\to(y, - x)\), substitute \( x = - 4\) and \( y=-2\). We get \((-2,4)\) (since \( - x=-(-4) = 4\)).
- For point \( R(-1, -2) \):
Substitute \( x=-1\) and \( y = - 2\) into the rule. We get \((-2,1)\) (since \( - x=-(-1)=1\)).
- For point \( S(-1, -4) \):
Substitute \( x = - 1\) and \( y=-4\) into the rule. We get \((-4,1)\) (since \( - x=-(-1) = 1\)). Wait, no, wait the rule is \((x,y)\to(y, - x)\). So for \( S(-1,-4) \), \( x=-1\), \( y = - 4\), so the new point is \((-4,1)\)? Wait, no, wait: \( (x,y)\to(y, - x) \). So \( x=-1\), \( -x = 1\), \( y=-4\). So the new point is \((-4,1)\)? Wait, no, I think I made a mistake. Wait, the correct rule for 90° clockwise rotation about the origin is \((x,y)\to(y, - x)\). Let's take an example: a point \((1,0)\) rotated 90° clockwise becomes \((0, - 1)\)? No, wait no. Wait, the standard rotation matrices: A 90° clockwise rotation matrix is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\). So if we have a vector \(\begin{pmatrix}x\\y\end{pmatrix}\), after rotation, it becomes \(\begin{pmatrix}y\\-x\end{pmatrix}\). So for a point \((x,y)\), the new point is \((y, - x)\).

Wait, let's take a point \((2,3)\). Rotated 90° clockwise: the new point is \((3, - 2)\). Let's verify with the rotation matrix: \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}2\\3\end{pmatrix}=\begin{pmatrix}3\\-2\end{pmatrix}\), which is correct.

So going back to our points:

- \( Q(-4, -2) \): \( x=-4\), \( y = - 2\). New point: \((y, - x)=(-2,4)\) (since \( - x=-(-4)=4\))
- \( R(-1, -2) \): \( x=-1\), \( y=-2\). New point: \((-2,1)\) (since \( - x=-(-1)=1\))
- \( S(-1, -4) \): \( x=-1\), \( y=-4\). New point: \((-4,1)\) (Wait, no: \( y=-4\), \( - x = 1\), so \((-4,1)\)? Wait, no, \((y, - x)\) means the x - coordinate of the new point is \( y\) and the y - coordinate is \( - x\). So for \( S(-1,-4) \), \( y=-4\), \( - x = 1\), so new point is \((-4,1)\)? Wait, that seems odd. Wait, maybe I misread the coordinates. Let's re - check the original figure.

Looking at the graph again: The figure \( QRST \) is a rectangle. Let's find the correct coordinates:

- \( Q \): Let's count the grid squares. From the origin (0,0), moving left 4 units (x=-4) and down 2 units (y=-2)? No, wait the y - axis: the positive y is up, negative y is down. The points \( Q\), \( R\), \( S\), \( T \):

\( Q \): x=-4, y=-2? \( R \): x=-1, y=-2? \( S \): x=-1, y=-4? \( T \): x=-4, y=-4? Yes, that makes a rectangle with length 3 (from x=-4 to x=-1) and height 2 (from y=-2 to y=-4).

Now, applying the 90° clockwise rotation rule \((x,y)\to(y, - x)\):

- \( Q(-4, -2) \): new coordinates \((-2,4)\) (because \( y=-2\), \( - x = 4\))
- \( R(-1, -2) \): new coordinates \((-2,1)\) (because \( y=-2\), \( - x = 1\))
- \( S(-1, -4) \): new coordinates \((-4,1)\) (because \( y=-4\), \( - x = 1\))
- \( T(-4, -4) \): new coordinates \((-4,4)\) (because \( y=-4\), \( - x = 4\))

Wait, no, that can't be right. Wait, maybe the rule is different. Wait, another way: 90° clockwise rotation about the origin: \((x,y)\) becomes \((y, - x)\). Let's take a point in the third quadrant (where our original points are, since x and y are negative). Rotating 90° clockwise should move it to the second quadrant (x negative, y positive) or first? Wait, no:

A point \((-a,-b)\) (a>0, b>0) rotated 90° clockwise: using the rule \((x,y)\to(y, - x)\), we get \((-b,a)\). Ah! I made a mistake earlier. The rule is \((x,y)\to(y, - x)\), so for \( x=-a\), \( - x=a\), and \( y=-b\), so the new point is \((-b,a)\).

So let's recalculate:

- For \( Q(-4, -2) \): \( x=-4\), \( y=-2\). New point: \((y, - x)=(-2,4)\) (since \( - x = 4\)) – wait, that's the same as before. Wait, \((-4,-2)\): \( x=-4\), \( y=-2\). So \( y=-2\), \( - x = 4\), so \((-2,4)\).

- For \( R(-1, -2) \): \( x=-1\), \( y=-2\). New point: \((-2,1)\) (since \( - x = 1\))

- For \( S(-1, -4) \): \( x=-1\), \( y=-4\). New point: \((-4,1)\) (since \( - x = 1\))

- For \( T(-4, -4) \): \( x=-4\), \( y=-4\). New point: \((-4,4)\) (since \( - x = 4\))

Now, let's plot these points:

- \( Q'(-2,4) \)
- \( R'(-2,1) \)
- \( S'(-4,1) \)
- \( T'(-4,4) \)

Wait, no, that seems to form a rectangle, but let's check with a different approach. The 90° clockwise rotation about the origin can also be thought of as a 270° counter - clockwise rotation. The rule for 90° clockwise rotation is indeed \((x,y)\to(y, - x)\).

Alternatively, let's use the rotation matrix. The rotation matrix for 90° clockwise is \( \begin{pmatrix}0&1\\-1&0\end{pmatrix} \).

For point \( Q(-4,-2) \):

\( \begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}-4\\-2\end{pmatrix}=\begin{pmatrix}-2\\4\end{pmatrix} \), so \( Q'(-2,4) \)

For point \( R(-1,-2) \):

\( \begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}-1\\-2\end{pmatrix}=\begin{pmatrix}-2\\1\end{pmatrix} \), so \( R'(-2,1) \)

For point \( S(-1,-4) \):

\( \begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}-1\\-4\end{pmatrix}=\begin{pmatrix}-4\\1\end{pmatrix} \), so \( S'(-4,1) \)

For point \( T(-4,-4) \):

\( \begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}-4\\-4\end{pmatrix}=\begin{pmatrix}-4\\4\end{pmatrix} \), so \( T'(-4,4) \)

Now, to draw the rotated figure, we connect \( Q'(-2,4) \), \( R'(-2,1) \), \( S'(-4,1) \), \( T'(-4,4) \) in order.

Answer:

The coordinates of the rotated figure \( Q'R'S'T' \) are \( Q'(-2,4) \), \( R'(-2,1) \), \( S'(-4,1) \), \( T'(-4,4) \). When plotted on the coordinate plane and connected, these points form the 90° clockwise rotation of \( QRST \) about the origin.