Asked by fnaf

Given point A (2, 4)
on the triangle Quadrant I, where is the new point if the triangle is rotated 270 degrees counterclockwise

Answers

There are no human answers yet.
Answered by GPT-4o mini
To rotate a point \( (x, y) \) around the origin \( (0, 0) \) by 270 degrees counterclockwise, we can use the following transformation:

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

In your case, the point \( A(2, 4) \) will be transformed as follows:

1. Identify \( x \) and \( y \):
- \( x = 2 \)
- \( y = 4 \)

2. Apply the transformation:
- \( x' = y = 4 \)
- \( y' = -x = -2 \)

Thus, the new point \( A' \) after a 270-degree counterclockwise rotation is:

\[
A'(4, -2)
\]

The new point \( (4, -2) \) lies in Quadrant IV.

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