Given point A (2, 4)

Question

on the triangle in Quadrant I, where is the new point if the triangle is rotated 90 degrees counterclockwise?

Answer 1

To rotate a point counterclockwise by 90 degrees around the origin, you can use the following transformation for a point \((x, y)\):

\[ (x', y') = (-y, x) \]

Given point \(A(2, 4)\):

1. The x-coordinate is \(2\) and the y-coordinate is \(4\).
2. Applying the transformation:
   • The new x-coordinate \(x' = -y = -4\)
   • The new y-coordinate \(y' = x = 2\)

Thus, the new point after rotating \(A(2, 4)\) by 90 degrees counterclockwise is:

\[ A'(-4, 2) \]

Point \(A'(-4, 2)\) is located in Quadrant II, as the x-coordinate is negative and the y-coordinate is positive.