Question

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

Explanation:

Step1: Recall rotation rules

For a point \((x,y)\) rotated \(270^\circ\) clockwise about the origin, the rule is \((x,y)\to(y, -x)\). But here we rotate about \((5,1)\), so first translate the point so that the center \((5,1)\) becomes the origin.

Step2: Translate point Q

Point \(Q\) is \((4, -7)\). To translate, subtract the center's coordinates: \(x' = 4 - 5=-1\), \(y' = -7 - 1=-8\). So the translated point is \((-1, -8)\).

Step3: Apply \(270^\circ\) clockwise rotation

Using the rule \((x',y')\to(y', -x')\), we get \(x''=-8\), \(y''=-(-1)=1\). So the rotated translated point is \((-8, 1)\).

Step4: Translate back

Add the center's coordinates: \(x = -8 + 5=-3\), \(y = 1 + 1=2\). Wait, no, wait. Wait, maybe I made a mistake. Wait, let's re - do the translation. Wait, when rotating about \((h,k)\), the steps are:
1. Subtract \((h,k)\) from the point: \((x - h,y - k)\)
2. Rotate the new point.
3. Add \((h,k)\) back to the rotated point.

Wait, let's re - calculate:

Original point \(Q=(4,-7)\), center \((h,k)=(5,1)\)

Step 1: Translate: \((4 - 5,-7 - 1)=(-1,-8)\)

Step 2: Rotate \(270^\circ\) clockwise. The rule for \(270^\circ\) clockwise rotation about the origin is \((x,y)\to(y, -x)\). So for \((-1,-8)\), applying the rule: \(x_{rot}= - 8\), \(y_{rot}=-(-1)=1\). So the rotated point is \((-8,1)\)

Step 3: Translate back: add \((h,k)=(5,1)\) to \((-8,1)\): \(x=-8 + 5=-3\), \(y = 1+1 = 2\)? Wait, that seems wrong. Wait, maybe I mixed up the rotation direction. Wait, \(270^\circ\) clockwise is the same as \(90^\circ\) counter - clockwise. The rule for \(90^\circ\) counter - clockwise is \((x,y)\to(-y,x)\). Wait, no, let's confirm the rotation rules:

- \(90^\circ\) clockwise: \((x,y)\to(y,-x)\)
- \(180^\circ\) clockwise: \((x,y)\to(-x,-y)\)
- \(270^\circ\) clockwise: \((x,y)\to(-y,x)\)

Ah! I made a mistake in the rotation rule. So for \(270^\circ\) clockwise, the rule is \((x,y)\to(-y,x)\)

So let's re - do step 2:

For the translated point \((-1,-8)\), applying \(270^\circ\) clockwise rotation: \(x_{rot}=-(-8)=8\), \(y_{rot}=-1\). So the rotated translated point is \((8,-1)\)

Step 3: Translate back: add \((5,1)\) to \((8,-1)\): \(x = 8+5 = 13\), \(y=-1 + 1=0\). No, that's not right. Wait, I think my initial approach to the rotation rule was wrong.

Wait, let's use another method. Let's consider the vector from the center to the point. The vector from \((5,1)\) to \((4,-7)\) is \(\vec{v}=(4 - 5,-7 - 1)=(-1,-8)\)

A \(270^\circ\) clockwise rotation of a vector \((a,b)\) is equivalent to a \(90^\circ\) counter - clockwise rotation. The rotation matrix for \(90^\circ\) counter - clockwise is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\). So if we have a vector \(\begin{pmatrix}a\\b\end{pmatrix}\), after \(90^\circ\) counter - clockwise rotation, it becomes \(\begin{pmatrix}-b\\a\end{pmatrix}\)

So for \(\vec{v}=\begin{pmatrix}-1\\-8\end{pmatrix}\), after rotation, the vector is \(\begin{pmatrix}-(-8)\\-1\end{pmatrix}=\begin{pmatrix}8\\-1\end{pmatrix}\)

Then the new point is the center plus the rotated vector: \((5 + 8,1+(-1))=(13,0)\). Wait, this is different from before. I must have messed up the rotation rule.

Wait, let's check the rotation rules again:

- \(90^\circ\) clockwise: \((x,y)\to(y,-x)\)
- \(180^\circ\) clockwise: \((x,y)\to(-x,-y)\)
- \(270^\circ\) clockwise: \((x,y)\to(-y,x)\)

Let's take a simple point, say \((1,0)\), rotate \(270^\circ\) clockwise about the origin. A \(270^\circ\) clockwise rotation of \((1,0)\) should end up at \((0, - 1)\)? Wait, no. Wait, if you rotate the point \((1,0)\) (on the x - axis) \(270^\circ\) clockwise, it will move to the negative y - axis. Wait, \((1,0)\) rotated \(90^\circ\) clockwise is \((0,-1)\), \(180^\circ\) clockwise is \((-1,0)\), \(270^\circ\) clockwise is \((0,1)\). Wait, so my previous rule was wrong.

Let's use the standard rotation matrices. The rotation matrix for an angle \(\theta\) counter - clockwise is \(\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\)

For \(270^\circ\) clockwise, \(\theta=- 270^\circ\) (since clockwise is negative angle). \(\cos(-270^\circ)=0\), \(\sin(-270^\circ)=1\)

So the rotation matrix is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) (because \(\cos(-270)=\cos(270) = 0\), \(\sin(-270)=-\sin(270)=1\))

So for a point \((x,y)\), the rotated point is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-y\\x\end{pmatrix}\)

Wait, when we rotate \((1,0)\) with this matrix: \(\begin{pmatrix}-0\\1\end{pmatrix}=(0,1)\), which is correct (rotating \((1,0)\) \(270^\circ\) clockwise gives \((0,1)\))

When we rotate \((0,1)\) with this matrix: \(\begin{pmatrix}-1\\0\end{pmatrix}=(-1,0)\)

When we rotate \((-1,0)\) with this matrix: \(\begin{pmatrix}-0\\-1\end{pmatrix}=(0,-1)\)

When we rotate \((0,-1)\) with this matrix: \(\begin{pmatrix}1\\0\end{pmatrix}=(1,0)\)

So the correct rule for \(270^\circ\) clockwise rotation about the origin is \((x,y)\to(-y,x)\)

Now, let's re - do the rotation about \((5,1)\):

1. Translate the point \(Q=(4,-7)\) to the center \((5,1)\): \((4 - 5,-7 - 1)=(-1,-8)\)

2. Rotate the translated point \((-1,-8)\) \(270^\circ\) clockwise about the origin. Using the rule \((x,y)\to(-y,x)\), we get \((-(-8),-1)=(8,-1)\)

3. Translate back by adding the center \((5,1)\) to the rotated point: \((8 + 5,-1+1)=(13,0)\)

Wait, that still seems odd. Wait, maybe the problem is that I misread the point. Wait, the point is \(Q=(4,-7)\), center \((5,1)\). Let's use another approach. Let's find the distance from \(Q\) to the center. The horizontal distance is \(4 - 5=-1\), vertical distance is \(-7 - 1=-8\). After rotating \(270^\circ\) clockwise, the horizontal and vertical distances will change. A \(270^\circ\) clockwise rotation is equivalent to a \(90^\circ\) counter - clockwise rotation. In a \(90^\circ\) counter - clockwise rotation, the horizontal distance becomes the negative of the vertical distance, and the vertical distance becomes the horizontal distance. Wait, the vector from center to \(Q\) is \(\vec{v}=(-1,-8)\). After \(90^\circ\) counter - clockwise rotation, the vector becomes \((8,-1)\) (because for a vector \((a,b)\), \(90^\circ\) counter - clockwise rotation gives \((-b,a)\), so \((-(-8),-1)=(8,-1)\)). Then the new point is center plus the new vector: \((5 + 8,1+(-1))=(13,0)\)

Wait, but let's check with a simple case. Suppose we have a point \((h + a,k + b)\) rotated \(270^\circ\) clockwise about \((h,k)\). Let's take \(h = 0,k = 0\) (origin), \(a = 1,b = 0\). Then the point is \((1,0)\), rotated \(270^\circ\) clockwise about origin is \((0,1)\), which matches the rule \((x,y)\to(-y,x)\) (since \((1,0)\to(0,1)\) when we do \(-y = 0,x = 1\)? Wait, no, \((1,0)\) with rule \((-y,x)\) is \((0,1)\), which is correct.

Another example: point \((0,1)\) about origin, rotated \(270^\circ\) clockwise: \((-1,0)\), which is \((-y,x)\) where \(x = 0,y = 1\), so \(-y=-1,x = 0\), correct.

So going back, the calculation seems correct. So the image of \(Q\) after rotating \(270^\circ\) clockwise about \((5,1)\) is \((13,0)\)

Wait, maybe I made a mistake in the rotation rule. Let's check the \(270^\circ\) clockwise rotation rule again. The rotation of a point \((x,y)\) \(270^\circ\) clockwise about the origin can also be thought of as rotating \(90^\circ\) counter - clockwise. The formula for \(90^\circ\) counter - clockwise rotation is \((x,y)\to(-y,x)\), and for \(270^\circ\) clockwise, it's the same as \(90^\circ\) counter - clockwise. So the rule is correct.

So after all the steps, the coordinates of the image should be \((13,0)\)

Answer:

\((13,0)\)