Question

graph the image of s(-7, -1) after a rotation 270° clockwise around the origin.

Explanation:

Step1: Recall rotation rule

The rule for a \(270^\circ\) clockwise rotation about the origin is \((x,y)\to(-y,x)\).

Step2: Apply the rule to \(S(-7,-1)\)

For the point \(S(-7,-1)\), here \(x = - 7\) and \(y=-1\).
Substitute into the rule: \(-y=-(-1) = 1\) and \(x=-7\) (wait, no, the rule is \((x,y)\to(-y,x)\), so new \(x\) is \(-y\) and new \(y\) is \(x\)).
So new \(x=-(-1)=1\), new \(y = - 7\)? Wait, no, wait: the correct rule for \(270^\circ\) clockwise (or \(90^\circ\) counter - clockwise) is \((x,y)\to(-y,x)\). Wait, let's re - derive it. A \(90^\circ\) counter - clockwise rotation is \((x,y)\to(-y,x)\), and a \(270^\circ\) clockwise rotation is the same as a \(90^\circ\) counter - clockwise rotation. Wait, no: \(270^\circ\) clockwise is equivalent to \(90^\circ\) counter - clockwise? Wait, no. Let's think about the rotation matrix. The rotation matrix for a \(270^\circ\) clockwise rotation (or \(- 270^\circ\) counter - clockwise) is \(

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

\). So if we have a point \((x,y)\), the new point \((x',y')\) is given by \(x'=y\) and \(y'=-x\)? Wait, I think I made a mistake earlier. Let's correct the rotation rule.

The correct rule for a \(270^\circ\) clockwise rotation about the origin: If we have a point \((x,y)\), after a \(270^\circ\) clockwise rotation, the new coordinates \((x',y')\) are given by \(x'=-y\) and \(y' = x\)? No, let's take an example. Let's take the point \((1,0)\). A \(270^\circ\) clockwise rotation: rotating \((1,0)\) \(270^\circ\) clockwise around the origin. A \(90^\circ\) clockwise rotation of \((1,0)\) is \((0, - 1)\), \(180^\circ\) clockwise is \((-1,0)\), \(270^\circ\) clockwise is \((0,1)\). Using the formula \((x,y)\to(-y,x)\): for \((1,0)\), \(-y = 0\), \(x = 1\), which gives \((0,1)\), which is correct. Another example: \((0,1)\) rotated \(270^\circ\) clockwise. Using the formula \((x,y)\to(-y,x)\): \(-y=-1\), \(x = 0\), so \((-1,0)\), which is correct (since \(270^\circ\) clockwise rotation of \((0,1)\) is \((-1,0)\)). Another example: \((1,1)\) rotated \(270^\circ\) clockwise. Using the formula, \(-y=-1\), \(x = 1\), so \((-1,1)\)? Wait, no, let's do it geometrically. Rotating \((1,1)\) \(270^\circ\) clockwise. The angle of the vector from the origin to \((1,1)\) is \(45^\circ\) from the x - axis. Rotating \(270^\circ\) clockwise, the new angle is \(45^\circ-270^\circ=- 225^\circ\) or \(135^\circ\) (since \(-225^\circ + 360^\circ=135^\circ\)). The magnitude is \(\sqrt{1 + 1}=\sqrt{2}\). The coordinates of a point with angle \(135^\circ\) and magnitude \(\sqrt{2}\) are \(\sqrt{2}\cos(135^\circ),\sqrt{2}\sin(135^\circ)\). \(\cos(135^\circ)=-\frac{\sqrt{2}}{2}\), \(\sin(135^\circ)=\frac{\sqrt{2}}{2}\). So the coordinates are \(\sqrt{2}\times(-\frac{\sqrt{2}}{2})=- 1\), \(\sqrt{2}\times\frac{\sqrt{2}}{2}=1\). So \((-1,1)\), which matches the formula \((x,y)\to(-y,x)\) (since \(x = 1\), \(y = 1\), \(-y=-1\), \(x = 1\)? Wait, no, \((-y,x)=(-1,1)\), yes. So the formula is correct: \((x,y)\to(-y,x)\) for \(270^\circ\) clockwise rotation.

Now, for the point \(S(-7,-1)\), \(x=-7\), \(y = - 1\).
Applying the rule \((x,y)\to(-y,x)\):
New \(x=-y=-(-1)=1\)
New \(y=x=-7\)
So the image of \(S(-7,-1)\) after a \(270^\circ\) clockwise rotation about the origin is \((1,-7)\).

Answer:

The image of \(S(-7,-1)\) after a \(270^\circ\) clockwise rotation around the origin is the point \((1, - 7)\). To graph it, we find the point with \(x = 1\) and \(y=-7\) on the coordinate plane.