Given point A (2, 6 on the triangle, where would the new point on the triangle be after a rotation of 90 degrees clockwise

Answer 1

To find the new coordinates after a rotation of 90 degrees clockwise, we can use the rotation matrix:

\[\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}\]

Here, θ (theta) is the angle of rotation.

In this case, we have a rotation of 90 degrees clockwise, so θ = -90 degrees (or -π/2 radians).

The coordinates of point A are (2,6), so we can substitute these values into the formula:

\[\begin{bmatrix} \cos(-90^\circ) & -\sin(-90^\circ) \\ \sin(-90^\circ) & \cos(-90^\circ) \end{bmatrix} \begin{bmatrix} 2 \\ 6 \end{bmatrix}\]

Using the trigonometric values of -90 degrees, we have:

\[\begin{bmatrix} \cos(-90^\circ) & -\sin(-90^\circ) \\ \sin(-90^\circ) & \cos(-90^\circ) \end{bmatrix} \begin{bmatrix} 2 \\ 6 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 2 \\ 6 \end{bmatrix}\]

Carrying out the matrix multiplication:

\[\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 2 \\ 6 \end{bmatrix} = \begin{bmatrix} (0 \cdot 2) + (1 \cdot 6) \\ (-1 \cdot 2) + (0 \cdot 6) \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix}\]

Therefore, the new coordinates after a rotation of 90 degrees clockwise would be (6,-2).