Question

1. point q with coordinate (4, -7) is rotated 270° clockwise about (5, 1). what are the coordinates of its image?

Explanation:

Step1: Translate to origin

First, translate the point \( Q(4, -7) \) and the center of rotation \( (5, 1) \) so that the center is at the origin. The translation vector is \( (-5, -1) \). So the translated coordinates of \( Q \) are \( (4 - 5, -7 - 1)=(-1, -8) \).

Step2: Rotate 270° clockwise

The rule for a 270° clockwise rotation about the origin is \( (x, y) \to (y, -x) \). Applying this to \( (-1, -8) \), we get \( (-8, 1) \) (since \( x=-1, y = -8 \); new \( x = y=-8 \), new \( y=-x = -(-1)=1 \)).

Step3: Translate back

Now, translate the rotated point back using the inverse translation vector \( (5, 1) \). So the new coordinates are \( (-8 + 5, 1 + 1)=(-3, 2) \)? Wait, no, wait. Wait, 270° clockwise is equivalent to 90° counterclockwise. Wait, maybe I mixed up the rule. Let's recall: 90° clockwise: \( (x,y)\to(y, -x) \); 180° clockwise: \( (x,y)\to(-x, -y) \); 270° clockwise: \( (x,y)\to(-y, x) \). Oh! I made a mistake earlier. Let's correct Step 2.

Correct Step 2: For 270° clockwise rotation, the rule is \( (x, y) \to (-y, x) \). So applying to \( (-1, -8) \): \( -y = -(-8)=8 \), \( x=-1 \). So the rotated point is \( (8, -1) \).

Now Step 3: Translate back by adding \( (5, 1) \) to \( (8, -1) \). So \( 8 + 5 = 13 \), \( -1 + 1 = 0 \)? Wait, no, wait. Wait, the translation to origin was \( (x - h, y - k) \) where \( (h,k)=(5,1) \). So to translate back, we do \( (x' + h, y' + k) \), where \( (x', y') \) is the rotated point in the translated system.

Wait, let's re - do Step 1: Original point \( Q(4,-7) \), center \( C(5,1) \). Vector \( \overrightarrow{CQ}=(4 - 5, -7 - 1)=(-1, -8) \).

270° clockwise rotation of a vector \( (a,b) \) about the origin: the rotation matrix for 270° clockwise is \( \begin{pmatrix}0&1\\-1&0\end{pmatrix} \) (wait, no: rotation matrix for \( \theta \) clockwise is \( \begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix} \). For \( \theta = 270^{\circ} \), \( \cos270^{\circ}=0 \), \( \sin270^{\circ}=-1 \). So the matrix is \( \begin{pmatrix}0&-1\\1&0\end{pmatrix} \)? Wait, maybe better to use the standard rules.

Alternative rule: 270° clockwise rotation: \( (x,y) \to (y, -x) \) is wrong. Let's use the fact that 270° clockwise is the same as 90° counter - clockwise. The rule for 90° counter - clockwise is \( (x,y)\to(-y, x) \). So for a point \( (x,y) \), 270° clockwise rotation: \( (x,y)\to(-y, x) \).

So for the translated point \( (-1, -8) \): \( -y=-(-8) = 8 \), \( x=-1 \). So the rotated point is \( (8, -1) \)? No, wait, 90° counter - clockwise: \( (x,y)\to(-y, x) \). So \( x=-1,y = -8 \): \( -y = 8 \), \( x=-1 \)? No, \( -y=8 \), \( x=-1 \), so \( (-y,x)=(8, -1) \)? Wait, no, \( x=-1,y=-8 \), so \( -y = 8 \), \( x=-1 \), so \( (8, -1) \). Then translate back: \( 8+5 = 13 \), \( -1 + 1=0 \). That can't be right. Wait, maybe my translation is wrong.

Wait, let's start over. The vector from the center \( (5,1) \) to \( Q(4,-7) \) is \( \vec{v}=(4 - 5, -7 - 1)=(-1, -8) \). We need to rotate this vector 270° clockwise. The rotation matrix for 270° clockwise is \( R=\begin{pmatrix}0&1\\-1&0\end{pmatrix} \) (wait, no: the rotation matrix for \( \theta \) degrees clockwise is \( \begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix} \). For \( \theta = 270^{\circ} \), \( \cos270^{\circ}=0 \), \( \sin270^{\circ}=-1 \). So \( R=\begin{pmatrix}0&-1\\1&0\end{pmatrix} \). Let's apply \( R \) to \( \vec{v}=(-1, -8) \).

\( R\vec{v}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}-1\\-8\end{pmatrix}=\begin{pmatrix}0\times(-1)+(-1)\times(-8)\\1\times(-1)+0\times(-8)\end{pmatrix}=\begin{pmatrix}8\\-1\end{pmatrix} \). So the rotated vector is \( (8, -1) \). Now, to find the image of \( Q \), we add this rotated vector to the center \( (5,1) \). So \( (5 + 8,1+(-1))=(13,0) \)? That seems off. Wait, maybe I messed up the rotation matrix.

Wait, let's use a different approach. Let's consider the rotation of a point \( (x,y) \) about \( (h,k) \) by \( \theta \) degrees. The formula is:

1. Translate the system: \( x'=x - h \), \( y'=y - k \)
2. Rotate \( (x',y') \) by \( \theta \) degrees: \( x''=x'\cos\theta - y'\sin\theta \), \( y''=x'\sin\theta + y'\cos\theta \)
3. Translate back: \( x=x''+h \), \( y=y''+k \)

For \( \theta = 270^{\circ} \), \( \cos270^{\circ}=0 \), \( \sin270^{\circ}=-1 \).

So \( x'=-1 \), \( y'=-8 \)

\( x''=(-1)\times0-(-8)\times(-1)=0 - 8=-8 \)

\( y''=(-1)\times(-1)+(-8)\times0 = 1+0 = 1 \)

Now translate back: \( x=-8 + 5=-3 \), \( y=1 + 1=2 \). Wait, now I get \( (-3,2) \). But why the discrepancy? Because the rotation matrix for 270° clockwise: when \( \theta = 270^{\circ} \), \( \cos\theta = 0 \), \( \sin\theta=-1 \). So the rotation formulas:

\( x''=x'\cos\theta - y'\sin\theta=x'\times0 - y'\times(-1)=y' \)

\( y''=x'\sin\theta + y'\cos\theta=x'\times(-1)+y'\times0=-x' \)

Ah! So I had the rotation rule wrong earlier. The correct rule for 270° clockwise rotation about the origin is \( (x,y)\to(y, -x) \). Let's verify with the matrix:

\( x''=y' \), \( y''=-x' \)

So for \( (x',y')=(-1, -8) \), \( x''=-8 \), \( y''=-(-1)=1 \). So the rotated point is \( (-8,1) \). Then translate back: \( x=-8 + 5=-3 \), \( y=1 + 1=2 \). Yes, that matches the matrix calculation. So my initial rotation rule was wrong. The correct rule for 270° clockwise is \( (x,y)\to(y, -x) \)? Wait, no, in the matrix, \( x''=y' \), \( y''=-x' \), so \( (x',y')\to(y', -x') \). So for \( (-1, -8) \), \( y'=-8 \), \( -x'=1 \), so \( (-8,1) \). Then translate back: \( -8 + 5=-3 \), \( 1+1 = 2 \). So the image is \( (-3,2) \).

Wait, let's check with an example. Take a point \( (0,1) \), rotate 270° clockwise about origin. 270° clockwise from (0,1) should be (1,0)? Wait, no. Wait, (0,1) is on the positive y - axis. Rotating 270° clockwise (which is three - quarters of a circle clockwise) would move it to the positive x - axis. So (0,1) rotated 270° clockwise is (1,0). Using the rule \( (x,y)\to(y, -x) \): \( (0,1)\to(1,0) \), which is correct. Another example: (1,0) rotated 270° clockwise: \( (0, -1) \). Using the rule: \( (1,0)\to(0, -1) \), correct. So the rule is \( (x,y)\to(y, -x) \) for 270° clockwise.

So going back, translated point \( (-1, -8) \): \( y'=-8 \), \( -x'=-(-1)=1 \), so \( (y', -x')=(-8,1) \). Then translate back: \( -8 + 5=-3 \), \( 1+1 = 2 \). So the image of \( Q \) after rotation is \( (-3,2) \).

Answer:

The coordinates of the image are \(\boxed{(-3, 2)}\)