Question

1. pq is rotated 270° clockwise about the origin to produce pq. write the rule for the rotation. (x, y) → (□, )

Explanation:

Step1: Recall rotation rules

A \(270^\circ\) clockwise rotation about the origin is equivalent to a \(90^\circ\) counter - clockwise rotation. The rule for a \(270^\circ\) clockwise rotation about the origin is \((x,y)\to(y, - x)\). Let's verify with a point. For example, take point \(Q\) from the graph. Let's assume the coordinates of \(Q\) are \((1,6)\) (from the graph, since it's on \(x = 1,y = 6\)). After \(270^\circ\) clockwise rotation, \(Q'\) has coordinates \((6,- 1)\)? Wait, no, looking at the graph, \(Q'\) is at \((-6,1)\)? Wait, maybe I made a mistake. Wait, let's take point \(P\). Let's say \(P\) is at \((0,- 6)\) (from the graph, the orange point \(P\) is on \(y=-6,x = 0\)), and \(P'\) is at \((6,0)\). Wait, no, the blue point \(P'\) is at \((6,0)\). So if \(P=(0,-6)\), applying the rule \((x,y)\to(y,-x)\), we get \((-6,0)\)? No, that's not matching. Wait, maybe I mixed up clockwise and counter - clockwise. The rule for \(270^\circ\) clockwise rotation: \((x,y)\) rotated \(270^\circ\) clockwise about the origin becomes \((y,-x)\)? Wait, no, let's recall the standard rotation rules. The rotation of a point \((x,y)\) about the origin:

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

Wait, let's test with point \(Q\). Let's say \(Q=(1,6)\) (from the graph, \(x = 1,y = 6\)). After \(270^\circ\) clockwise rotation, \(Q'\) should be \((-6,1)\) (from the graph, \(Q'\) is at \((-6,1)\)). Let's apply the rule \((x,y)\to(-y,x)\). For \(Q=(1,6)\), \(-y=-6\), \(x = 1\), so \((-6,1)\), which matches. Now test point \(P\). Let's say \(P=(0,-6)\) (the orange point \(P\) is at \(x = 0,y=-6\)). Applying the rule \((x,y)\to(-y,x)\), we get \((6,0)\), and \(P'\) is at \((6,0)\) (the blue point \(P'\)), which matches. So the correct rule for \(270^\circ\) clockwise rotation about the origin is \((x,y)\to(-y,x)\).

Step2: Confirm with the graph points

Take point \(Q(1,6)\): apply \((x,y)\to(-y,x)\), we get \((-6,1)\), which is the coordinates of \(Q'\) on the graph. Take point \(P(0,-6)\): apply \((x,y)\to(-y,x)\), we get \((6,0)\), which is the coordinates of \(P'\) on the graph. So the rule holds.

Answer:

\((x, y)\to(-y, x)\)