Question

P(−7,2)
is rotated 270°
(counterclockwise) about the origin. Name the quadrant P′
is located in. Enter the option number that corresponds to your answer.

Option #1: Quadrant I

Option #2: Quadrant II

Option #3: Quadrant III

Option #4: Quadrant IV

(1 point)
P′
is located in Option #
.

Answer 1

To find the new coordinates of the point \( P(-7,2) \) after a 270° counterclockwise rotation about the origin, we can use the rotation formula for points in the Cartesian coordinate system:

For a point \( (x,y) \) rotated counterclockwise by an angle \( \theta \), the new coordinates \( (x', y') \) are given by:
\[
x' = x \cdot \cos(\theta) - y \cdot \sin(\theta)
\]
\[
y' = x \cdot \sin(\theta) + y \cdot \cos(\theta)
\]

For a 270° counterclockwise rotation, we have:
\[
\cos(270^\circ) = 0, \quad \sin(270^\circ) = -1
\]

Using the coordinates of point \( P(-7, 2) \):

1. Calculate \( x' \):
\[
x' = -7 \cdot 0 - 2 \cdot (-1) = 0 + 2 = 2
\]

2. Calculate \( y' \):
\[
y' = -7 \cdot (-1) + 2 \cdot 0 = 7 + 0 = 7
\]

Therefore, after the rotation, the coordinates of the point \( P' \) are \( (2, 7) \).

Now, let's determine the quadrant where point \( P'(2, 7) \) is located:

- The x-coordinate (2) is positive.
- The y-coordinate (7) is positive.

Thus, point \( P'(2, 7) \) is in **Quadrant I**.

The answer is **Option #1: Quadrant I**.