Given point A (2, 4) on the triangle in Quadrant 1, where is the new point if the triangle is rotated 90 degrees counterclockwise???
To rotate a point counterclockwise by 90 degrees, we can use the rotation matrix:
\[ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \]
Multiplying the matrix by the coordinates of point A, we get:
\[ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \end{bmatrix} = \begin{bmatrix} -4 \\ 2 \end{bmatrix} \]
Therefore, the new point after rotating A by 90 degrees counterclockwise would be (-4, 2).