given A (2, 4) on the triangle in Quadrant I, how would you describe the coordinates of the new point when the triangle is rotated 90 degrees clockwise
To rotate a point 90 degrees clockwise, we can use the rotation matrix:
\[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \]
For a 90 degrees clockwise rotation, we have \(\theta = -90^\circ\) or \(\theta = -\frac{\pi}{2}\) in radians. Substituting the given coordinates \(x = 2\) and \(y = 4\), and applying the rotation formula, we get:
\[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos(-\frac{\pi}{2}) & -\sin(-\frac{\pi}{2}) \\ \sin(-\frac{\pi}{2}) & \cos(-\frac{\pi}{2}) \end{pmatrix} \begin{pmatrix} 2 \\ 4 \end{pmatrix} \]
Using the values of sine and cosine for \(-\frac{\pi}{2}\), our expression becomes:
\[ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 2 \\ 4 \end{pmatrix} \]
Multiplying the matrices, we get:
\[ \begin{pmatrix} 0 \cdot 2 + 1 \cdot 4 \\ -1 \cdot 2 + 0 \cdot 4 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} \]
Therefore, when the triangle is rotated 90 degrees clockwise, the new coordinates for the point A (2, 4) would be (4, -2).