Rotate △ABC (5,-1) (2,-7) (5,-9)

90 degrees (counterclockwise) about the origin to create △A′B′C′
. What are the coordinates of the following points?

1 answer

To rotate a point \((x, y)\) 90 degrees counterclockwise about the origin, the new coordinates will be \((-y, x)\).

Let's apply this to each of the points of triangle \( \triangle ABC \):

  1. For point \( A(5, -1) \): \[ A' = (-(-1), 5) = (1, 5) \]

  2. For point \( B(2, -7) \): \[ B' = (-(-7), 2) = (7, 2) \]

  3. For point \( C(5, -9) \): \[ C' = (-(-9), 5) = (9, 5) \]

Thus, the coordinates of the points after the rotation are:

  • \( A'(1, 5) \)
  • \( B'(7, 2) \)
  • \( C'(9, 5) \)

The new coordinates of triangle \( \triangle A'B'C' \) are \( A'(1, 5), B'(7, 2), C'(9, 5) \).