Given point A (2, 4) on the triangle Quadrant I, where is the new point if the triangle is rotated 270 degrees counterclockwise?
To rotate a point counterclockwise around the origin, we can use the following rotation formula:
(x', y') = (x cosθ - y sinθ, x sinθ + y cosθ)
where (x, y) are the coordinates of the original point, θ is the angle of rotation (in radians), and (x', y') are the coordinates of the rotated point.
In this case, the original point A is (2, 4) and we want to rotate it 270 degrees counterclockwise. To convert 270 degrees to radians, we multiply by π/180:
θ = 270 * π/180 = 3π/2
Using the rotation formula, we can find the new coordinates (x', y'):
(x', y') = (2 cos(3π/2) - 4 sin(3π/2), 2 sin(3π/2) + 4 cos(3π/2))
Since cos(3π/2) = 0 and sin(3π/2) = -1, we can simplify this as:
(x', y') = (0 - 4*(-1), 2*(-1) + 4*0)
= (0 + 4, -2 + 0)
= (4, -2)
Therefore, the new point after rotating 270 degrees counterclockwise is (4, -2).