Question

Given point A (2, 4)
𝐴

(
2
,

4
)
on the triangle Quadrant I, where is the new point if the triangle is rotated 270 degrees counterclockwise?

(1 point)
Responses

(−4, 2)
(
−
4
,

2
)
left parenthesis negative 4 comma 2 right parenthesis

(−2,−4)
(
−
2
,
−
4
)
left parenthesis negative 2 comma negative 4 right parenthesis

(4,−2)
(
4
,
−
2
)
left parenthesis 4 comma negative 2 right parenthesis

(−4,−2)

Answer 1

To find the new coordinates of point A (2, 4) after a 270-degree counterclockwise rotation, we can use the rotation transformation formula for a point (x, y) rotated by θ degrees around the origin (0, 0):

The new coordinates (x', y') after a rotation by θ degrees are given by:

\[
x' = x \cos(\theta) - y \sin(\theta)
\]

\[
y' = x \sin(\theta) + y \cos(\theta)
\]

For a 270-degree rotation (which is equivalent to a 90-degree clockwise rotation), we know:

- \(\cos(270^\circ) = 0\)
- \(\sin(270^\circ) = -1\)

Substituting these values into our formulas:

\[
x' = 2 \cdot 0 - 4 \cdot (-1) = 0 + 4 = 4
\]

\[
y' = 2 \cdot (-1) + 4 \cdot 0 = -2 + 0 = -2
\]

So after a 270-degree counterclockwise rotation, the new coordinates of point A (2, 4) will be (4, -2).

Thus, the answer is:

\[
\text{(4, -2)}
\]