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

Answer 1

To find the new point of the triangle after rotating 270 degrees counterclockwise, we can use the following rotation formula:

For counterclockwise rotation of (x, y) about the origin by angle θ, the new coordinates of the point (x', y') are:

x' = x * cos(θ) - y * sin(θ)
y' = x * sin(θ) + y * cos(θ)

In this case, since we are rotating 270 degrees counterclockwise, the angle θ is 270 degrees or 3π/2 radians.

Substituting the values of point A into the formula:

x' = 2 * cos(3π/2) - 4 * sin(3π/2)
y' = 2 * sin(3π/2) + 4 * cos(3π/2)

We know that cos(3π/2) = 0 and sin(3π/2) = -1, so plugging in these values:

x' = 2 * 0 - 4 * (-1) = 0 + 4 = 4
y' = 2 * (-1) + 4 * 0 = -2 + 0 = -2

Therefore, the new point after rotating 270 degrees counterclockwise is (4, -2).