QUESTION IMAGE
Question
- what will be the coordinates of point n after a 270° counterclockwise rotation about the origin? n (______)
Response
- Recall the rotation rule for a \(270^{\circ}\) counter - clockwise rotation about the origin:
- The rule for rotating a point \((x,y)\) \(270^{\circ}\) counter - clockwise about the origin is \((x,y)\to(y, - x)\).
- First, we need to determine the coordinates of point \(N\) from the graph. Assume the coordinates of point \(N\) are \((x_0,y_0)\). Let's say from the graph, if point \(N\) has coordinates \((a,b)\).
- Apply the rotation rule:
- After applying the \(270^{\circ}\) counter - clockwise rotation rule \((x,y)\to(y, - x)\) to point \(N(a,b)\), the new coordinates \(N'\) will be \((b,-a)\).
- For example, if \(N=(2,1)\), then after a \(270^{\circ}\) counter - clockwise rotation about the origin, \(N'=(1, - 2)\).
Since the coordinates of the original point \(N\) are not given in the question, we can only state the general rule. If we assume the coordinates of \(N\) are \((x,y)\), then:
Step1: Identify rotation rule
The rule for \(270^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(y, - x)\).
Step2: Apply rule to point \(N\)
If \(N=(x,y)\), then \(N'=(y, - x)\).
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If the original coordinates of \(N\) are \((x,y)\), the coordinates of \(N'\) are \((y,-x)\)