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

Answer 1

To find the new point after rotating point A (2, 4) 270 degrees counterclockwise, we can use the rotation matrix:

[x', y'] = [cos(theta) -sin(theta); sin(theta) cos(theta)] * [x, y]

Where theta is the angle of rotation (270 degrees counterclockwise in this case).

Using the rotation matrix:

[x', y'] = [cos(270) -sin(270); sin(270) cos(270)] * [2, 4]
= [0 -1; 1 0] * [2, 4]
= [0*(-2) + (-1)*(4); 1*(2) + 0*(4)]
= [-4; 2]

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