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